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ADVANCED  COURS 


SLAUGHT  &  LENNES 


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■iiliwwiimiwimiwiwuiMHnwmnmiiirtwiiiiiiiiiiiiHiiiiiiiiir  inr 


IN   MEMORIAM 

FLOR1AN  CAJOR1 


(T^w'cc^       C^^trv-t: 


HIGH  SCHOOL  ALGEBRA 

Hbvancefc  Course 


BY 


H.    E.    SLAUGHT,    Ph.D. 

ASSOCIATE   PROFESSOR   OF   MATHEMATICS    IN   THE   UNIVERSITY 
OF   CHICAGO 


N.    J.    LENNES,    Ph.D. 

INSTRUCTOR   IN    MATHEMATICS    IN   THE   MASSACHUSETTS 
INSTITUTE   OF   TECHNOLOGY 


JHHc 


Boston 

ALLYN    AND    BACON 

1908 


COPYRIGHT,  1908, 
BY  H.  E.  SLAUGHT 
AND  N.  J.  LENNES. 


m       M    Sm       m. 


PREFACE 

The  Advanced  Course  of  the  High  School  Algebra  contains 
a  review  of  all  topics  treated  in  the  Elementary  Course,  to- 
gether with  such  additional  topics  as  are  required  to  make  it 
amply  sufficient  to  meet  the  entrance  requirements  of  any 
college  or  technical  school.  Its  development  is  based  upon 
the  following  important  considerations : 

1.  The  pupil  has  had  a  one  year's  course  in  algebra,  involv- 
ing constant  application  of  its  elementary  processes  to  the 
solution  of  concrete  problems.  This  has  invested  the  pro- 
cesses themselves  with  an  interest  which  now  makes  them 
a  proper  object  of  study  for  their  own  sake. 

2.  The  pupil  has,  moreover,  developed  in  intellectual  ma- 
turity and  is,  therefore,  able  to  comprehend  processes  of 
reasoning  with  abstract  numbers  which  were  entirely  beyond 
his  reach  in  the  first  year's  course.  This  is  particularly  true 
if,  in  the  meantime,  he  has  learned  to  reason  with  the  more 
concrete  forms  of  geometry. 

In  consequence  of  these  considerations,  the  treatment 
throughout  is  from  a  more  mature  point  of  view  than  in  the 
Elementary  Course.  The  principles  of  algebra  are  given  in 
the  form  of  theorems  the  proofs  of  which  are  based  upon  a 
definite  set  of  axioms. 

As  in  the  Elementary  Course,  the  important  principles  are 
used  at  once  in  the  solution  of  concrete  and  interesting  prob- 
lems, which,  however,  are  here  adapted  to  the  pupil's  greater 
maturity  and  experience.  But  relatively  greater  space  and 
emphasis  are  given  to  the  manipulation  of  standard  algebraic 

iii 


msosisi 


IV 


PREFACE 


forms,  such  as  the  student  is  likely  to  meet  in  later  work  in 
mathematics  and  physics,  and  especially  such  as  were  too  com- 
plicated for  the   Elementary  Course. 

The  division  of  the  High  School  Algebra  into  two  distinct 
courses  has  made  it  possible  to  give  in  the  Advanced  Course 
a  more  thorough  treatment  of  the  elements  of  algebra  than 
could  be  given  if  the  book  were  designed  for  first-year  classes. 
It  has  thus  become  possible  to  lay  emphasis  upon  the  pedagogic 
importance  of  viewing  each  subject  a  second  time  in  a  manner 
more  profound  than  is  possible  on  a  first  view. 

Attention  is  specifically  called  to  the  following  points  : 

The  scientific  treatment  of  axioms  in  Chapter  I. 

The  clear  and  simple  treatment  of  equivalent  equations  in 
Chapter  III. 

The  discussion  by  formula,  as  well  as  by  graph,  of  incon- 
sistent and  dependent  systems  of  linear  equations,  pages  40 
to   I  I. 

The  unusually  complete  treatment  of  factoring  and  the 
clear  and  simple  exposition  of  the  general  process  of  finding 
the  Highest  Common  Factor,  in  Chapter  V. 

The  careful  discrimination  in  stating  and  applying  the 
theorems  on  powers  and  roots  in  Chapter  VI. 

The  unique  treatment  of  quadratic  equations  in  Chapter  VI  1, 
giving  a  lucid  exposition  in  concrete  and  graphical  form  of 
distinct,  coincident,  and  imaginary  roots. 

The  concise  treatment  of  radical  expressions  in  Chapter  X, 
and  especially  —  an  innovation  much  needed  in  this  connec- 
tion—  the  rich  collection  of  problems,  in  the  solution  of  which 
radicals  are  applied. 


H.  E.  SLAUGHT. 
N.   .1.   LENNES. 


<  '  II  I<    \iim     \M>    I'.OSTON, 

April,  L908. 


CONTENTS 


CHAPTER    I 
FUNDAMENTAL   LAWS 

Axioms  of  Addition  and'  Subtraction 
Axioms  of  Multiplication  and  Division 
Theorems  on  Addition  and  Subtraction 
Theorems  on  Multiplication  and  Division  . 


PAGE 

1 


CHAPTER    II 

FUNDAMENTAL   OPERATIONS 

Definitions 

Addition  and  Subtraction  of  Monomials  . 
Addition  and  Subtraction  of  Polynomials  . 
Removal  of  Parentheses  .... 
Multiplication  and  Division  of  Monomials 
Multiplication  and  Division  of  Polynomials 


11 
12 
15 
17 
18 
21 


CHAPTER   III 

INTEGRAL  EQUATIONS  OF   THE  FIRST  DEGREE  IN  ONE 
UNKNOWN 


Definitions 
Equivalent  Equations 
Problems  in  One  Unknown 


25 

27 
32 


CHAPTER  IV 

INTEGRAL   LINEAR  EQUATIONS  IN   TWO  OR  MORE  VARIABLES 
Indeterminate  Equations    ...         .....         .36 

Simultaneous  Equations  in  Two  Variables 38 


VI 


CONTENTS 


Inconsistent  and  Dependent  Equations 
Systems  in  .More  than  Two  Variables 
Problems  in  Two  or  More  Unknowns 

CHAPTER  V 

FACTORING 
Expression  of  Two,  Three,  or  Four  Terms 
Factors  found  by  Grouping 
Factors  found  by  the  Factor  Theorem 
Solution  of  Equations  by  Factoring   . 
Common  Factors  and  Multiples 

CHAPTER    VI 
POWERS   AND   ROOTS 
Definitions  ....... 

Theorems  on  Powers  and  Roots 

Roots  of  Polynomials  ..... 

Roots  of  Arabic  Numbers 


PAGE 

43 
45 
47 


52 

55 

■  u 

60 
Gl 


69 

73 

77 


CHAPTER   VII 

QUADRATIC  EQUATIONS 
Exposition  by  means  of  Graphs 
Distinct,  Coincident,  and  Imaginary  Roots 
Simultaneous  Quadratics    .... 
Special  Methods  of  Solution 
Higher  Equations  involving  Quadratics 
Relations  between  the  Roots  and  Coefficients 
Formation  of  Equations  with  Given  Roots 
Problems  involving  Quadratics  . 

CHAPTER    VIII 
ALGEBRAIC  FRACTIONS 
Reduction  of  Fractions        ..... 

Addition  and  Subtraction  of  Fractions 


83 
87 
90 
96 
106 
L08 

110 

112 


11G 

120 


CONTENTS 


Vll 


Multiplication  and  Division  of  Fractions    . 

Complex  Fractions 

Equations  involving  Algebraic  Fractions    .... 
Problems  involving  Fractions 

CHAPTER   IX 
RATIO,  VARIATION,   AND  PROPORTION 

Ratio  and  Variation 

Proportion  .......... 

Problems    .......... 


I'A(iE 

123 
126 

127 
133 


135 
139 
141 


CHAPTER   X 
EXPONENTS   AND   RADICALS 

Fractional  and  Negative  Exponents 142 

Reduction  of  Radical  Expressions 147 

Addition  and  Subtraction  of  Radicals 150 

Multiplication  and  Division  of  Radicals 151 

Rationalizing  the  Divisor    .         .         .         .         .         .         .         .         .155 

Square  Root  of  Radical  Expressions 156 

Equations  containing  Radicals  .         .         .         .         .         .         .         .159 

Problems  involving  Radicals 163 

CHAPTER   XI 

LOGARITHMS 

Definitions  and  Principles 167 

CHAPTER   XII 
PROGRESSIONS 


Arithmetic  Progressions 
Geometric  Progressions 
Harmonic  Progressions 


CHAPTER    XIII 
THE   BINOMIAL   FORMULA 


Proof  by  Induct  ion 
The  General  Term 


175 

181 
186 


189 
191 


HIGH   SCHOOL   ALGEBRA 

ADVANCED  COURSE 
CHAPTER  I 

FUNDAMENTAL   LAWS 

1.  We  have  seen  in  the  Elementary  Course  that  algebra,  like 
arithmetic,  deals  with  numbers  and  with  operations  upon  num- 
bers. We  now  proceed  to  study  in  greater  detail  the  laws  that 
underlie  these  operations. 

THE   AXIOMS   OF   ADDITION   AND   SUBTRACTION 

In  adding  numbers  we  assume  at  the  outset  certain  axioms. 

2.  Axiom  I.  Any  two  numbers  have  one  and  only  one 
sum. 

Since  two  numbers  are  equal  when  and  only  when  they  are 
the  same  number,  it  follows  from  this  axiom  that  if  a  =  b  and 
c  =  d  then  a  +  c  =  6  -f  d. 

For  if  a  is  the  same  number  as  b,  and  c  is  the  same  number  as  d, 
then  adding  b  and  d  is  the  same  as  adding  a  and  c,  and  by  Axiom  I 
the  sums  are  the  same  and  hence  equal. 

Therefore  from  Axiom  I  follows  the  axiom  usually  given  : 
If  equal  numbers  be  added  to  equal  numbers,  the  sums  are  equal 
numbers. 

Since  Axiom  I  asserts  that  the  sum  of  two  numbers  is  unique, 
it  is  often  called  the  uniqueness  axiom  of  addition. 

3.  If  a  =  c  and  b  =  c  then  a  =  b,  since  the  given  equations 
assert  that  a  is  the  same  number  as  b.  Hence  the  usual  state- 
ment: If  each  of  tiro  numbers  is  equal  to  the  same  number,  they 
are  equal  to  each  other. 

1 


2  FUNDAMENTAL  LAWS 

4.  The  sum  of  two  numbers,  as  G  and  8,  may  be  found  by 
ailding  6  to  8  or  8  to  6,  in  either  ease  obtaining  14  as  the 
result.  This  is  a  particular  case  of  a  general  law  for  all 
numbers  of  algebra,  which  we  enunciate  as 

Axiom  II.  The  smt)  of  two  numbers  is  the  same  in  what- 
ever order  they  are  added . 

This  is  expressed  in  symbols  by  the  identity  : 

a  +  b  =  b  +  a.  [See  §  37,  E.  C.*] 

Axiom  II  states  what  is  called  the  commutative  law  of  addition, 
since  it  asserts  that  numbers  to  be  added  may  be  commuted  or 
i .  1 1 1  i  rchanged  in  order. 

Definition.  Numbers  which  are  to  be  added  are  called  addends. 

5.  In  adding  three  numbers  such  as  5,  6,  and  7,  we  first  add 
two  of  them  and  then  add  the  third  to  this  sum.  It  is  imma- 
terial whether  we  first  add  5  and  6  and  then  add  7  to  the  sum, 
or  first  add  6  and  7  and  then  add  5  to  the  sum.  This  is  a  par- 
ticular case  of  a  general  law  for  all  numbers  of  algebra,  which 
we  enunciate  as 

Axiom  III.  The  sum  of  three  numbers  is  tiie  same  in 
whatever  man  iter  then  are  grouped. 

In  symbols  we  have 

a  +  b  +  c  =  a  +  (b  +  c). 

When  no  symbols  of  grouping  are  used,  we  understand  a  +  b  +  c 
to  mean  thai  a  and  b  are  to  be  added  first  and  then  <■  is  to  be  added 
to  the  sum. 

Axiom  III  states  what  is  called  the  associative  law  of  addi- 
tion, since  it  asserts  that  addends  may  be  associated  or  grouped 
in  any  desired  manner. 

It  is  to  be  noted  that  an  equality  may  be  read  in  either  direction. 
Thus        a  +  b  +  c=a+(b  +  c)  and  a  +  (b  +  c)=a  +  b  +  c 
are  equivalent  statements. 

6.  If  any  two  numbers,  such  as  1(.>  and  25,  are  given,  then 
in  arithmetic  we  can  always  find  a  number  which  added  to 

*  E.  C.  meaus  the  Elementary  Course. 


AXIOMS  3 

the  smaller  gives  the  larger  as  a  sum.     That  is,  we  can  sub- 
tract the  smaller  number  from  the  larger. 

In  Algebra,  where  negative  numbers  are  used,  any  number 
may  be  subtracted  from  any  other  number.     That  is : 

Axiom  IV.  For  any  pair-  of  numbers  a  and  b  there  is 
one  and  only  one  number  c  such  that  a  +  c  —  b. 

The  process  of  finding  the  number  c  when  a  and  b  are  given 
is  called  subtraction,  b  is  the  minuend,  a  the  subtrahend,  and  c 
the  remainder.     This  operation  is  also  indicated  thus,  b  —  a  =  c. 

If  a  +  c  =  a,  then  the  number  c  is  called  zero,  and  is  written 
0.     That  is,  a  +  0  =  a,  or  a  —  a  =  0. 

Adding  a  to  each  member  of  the  equality  b  —  a  =  c,  we  have 
b  —  a  +  a.=  c  +  a,  which  by  hypothesis  is  equal  to  b.  Hence 
subtracting  a  number  and  then  adding  the  same  member  gives  as 
a  result  the  original  number  operated  upon. 

Axiom  IV  is  called  the  uniqueness  axiom  of  subtraction.  A 
direct  consequence  is  the  following :  If  equal  numbers  are  sub- 
tracted from  equal  numbers,  the  remainders  are  equal  numbers. 

THE    AXIOMS    OF    MULTIPLICATION    AND    DIVISION 

7.  Axioms  similar  to  those  just  given  for  addition  and  sub- 
traction hold  for  multiplication  and  division. 

Axiom  V.    Two  numbers  have  one  and  only  one  product. 

This  is  called  the  uniqueness  axiom  of  multiplication.  It  is  a 
direct  consequence  of  this  axiom  that :  If  equal  numbers  are 
multiplied  by  equal  numbers,  the  products  are  equal  members. 

8.  The  product  of  5  and  6  may  be  obtained  by  taking  5  six 
times,  or  by  taking  6  five  times.  That  is,  5-6  =  6-5.  This 
is  a  special  case  of  a  general  law  for  all  numbers  of  algebra, 
which  we  enunciate  as 

Axiom  VI.    The  product  of  two  numbers  is  the  same  in 
whatever  order  they  are  multiplied. 
In  symbols  we  have  a  •  b  =  b  •  a. 


4  FUNDAMENTAL   LAWS 

This  axiom  states  what  is  called  the  commutative  law  of 
factors  in  multiplication. 

9.  The  product  of  three  numbers,  such  as  5,  6,  and  7,  may 
be  obtained  by  multiplying  5  and  6,  and  this  product  by  7,  or  6 
and  7,  and  this  product  by  5.  This  is  a  special  case  of  a 
general  law  for  all  numbers  of  algebra,  which  we  enunciate  as 

Axiom  VII.    The  product  of  three  numbers  is  the  same 
in  whatever  manner  they  are  grouped. 
In  symbols  we  have       abc  =  a(bc). 

The  expression  ahc  without  symbols  of  grouping  is  understood  to 
mean  that  the  product  of  a  and  b  is  to  be  multiplied  by  c. 

This  axiom  states  what  is  called  the  associative  law  of  factors 
in  multiplication. 

Principles  III  and  XV  of  E.  C.  follow  from  Axioms  VI  and  VII. 

10.  Another  law  for  all  numbers  of  algebra  is  enunciated  as 

Axiom  VIII.  The  product  of  the  sum  or  difference  of  tic<> 
numbers  and  a  given  number  is  equal  tothe  result  obtained 
by  multiplying  each  number  separately  by  the  given  num- 
ber and  then  adding  or  subtracting  theproducts. 

In  symbols  we  have 

a(b  -f  c)  =  ab  +  ac  and  a(6  —  c)  =  ab  —  ac. 

Axiom  VIII  states  what  is  called  the  distributive  law  of 
multiplication. 

When  these  identities  are  read  from  left  to  right,  they  are  equiva- 
lent to  Principle  IV.  E.  C,  and  when  rend  from  righl  to  left  (see  §  5) 
they  are  equivalent  to  Principles  I  and  II.  E.  C.  In  the  form 
a(b  ±  c)  =  ab  ±  ac  this  axiom  is  directly  applicable  to  the  multiplica- 
tion of  a  polynomial  by  a  monomial,  and  in  the  form  ab  ±  ac  =a(l>  ±  c), 
to  the  addition  and  subtraction  of  monomials  having  a  common  factor. 

11.  Axiom  IX.    For  any  two  numbers,  a  and  b.  provided 

a  is  not  equal  to  zero.'  there  is  one  and  only  one  number 
c  such  that  a  •  c  =  b. 

*The  symbol  for  the  expression  a  is  not  equal  to  zero  is  o  =f=  0. 


AXIOMS  5 

Definitions.  If  ac  =  b,  the  process  of  finding  c  when  a  and  b 
are  given  is  called  division,     b  is  the  dividend,  a  the  divisor, 

c  the  quotient,  and  we  write  b  -f-  a  =  c,  or  -  =  c.    For  the  case, 
a  =  0,  see  §§  24,  25. 

If  a  •  c  =  a,  a  =f=  0,  then  the  number  c  is  called  unity,  and  is 

written  1.     That  is,  -  =  1. 

a  7 

Multiplying  both  sides  of  the  equality  -  =  c  by  a,  we  have 

b  a 

a  •  -  =  ac,  which  by  hypothesis  equals  b.     Hence  dividing  by  a 
a 

number  and,  then  multiplying  by  the  same  number  gives  as  a  result 
the  original  number  operated  upon. 

Axiom  IX  is  called  the  uniqueness  axiom  of  division.  As  a 
direct  consequence  of  this  axiom  we  have :  If  equal  mem- 
bers are  divided  by  equal  numbers,  the  quotients  are  equal 
numbers. 

12.  Axioms  I,  IV  (incase  the  subtrahend  is  not  greater  than 
the  minuend),  V,  and  IX  underlie  respectively  the  processes 
of  addition,  subtraction,  multiplication,  and  division,  from  the 
very  beginning  in  elementary  arithmetic.  Axioms  II,  III,  VI, 
VII,  and  VIII  are  also  fundamental  in  arithmetic,  where  they 
are  usually  assumed  without  formal  statement. 

E.g.  Axiom  VIII  is  used  in  long  multiplication  such  as  125  x  235, 
where  we  multiply  125  by  5,  by  30,  and  by  200,  and  then  add  the 
products. 

13.  Negative  Numbers.  Axiom  IV,  in  case  the  subtrahend  is 
greater  than  the  minuend,  does  not  hold  in  arithmetic  because 
of  the  absence  of  the  negative  number.  This  axiom  therefore 
brings  the  negative  number  into  algebra. 

We  now  proceed  to  study  the  laws  of  operation  upon  this 
enlarged  number  .system.  In  the  Elementary  Course  concrete 
applications  were  used  to  show  that  certain  rules  of  signs  hold 
in  operations  upon  positive  and  negative  numbers.  We  shall 
now  see  that  the  same  rules  follow  from  the  axioms  just 
stated. 


6  FUNDAMENTAL  LAWS 

14.  Definitions.  If  a  +  b  =  0,  then  b  is  said  to  be  the  negative 
of  a  and  a  the  negative  of  b.  If  a  is  a  positive  number,  that 
is  an  ordinary  number  of  arithmetic,  then  h  is  called  a  negative 
number.  We  denote  the  negative  of  a  by  —  a.  Hence, 
a  +  (  —  a)  =  0.     </  and  —  a  have  the  same  absolute  value. 

If  a  —  b  is  positive,  then  a  is  said  to  be  greater  than  b. 
This  is  written  a  >  ?/.  If  a  —  b  is  negative,  then  a  is  said 
to  be  less  than  b.  This  is  written  a  <  b.  If  a  —  6  =  0,  then 
a  =  6,  and  if  a  =  6  then  a  —  6  =  0.     See  §  G. 

THEOREMS   ON   ADDITION   AND   SUBTRACTION 

Definition.     A  theorem  is  a  statement  to  be  proved. 
A  corollary  is  a  theorem  which  follows  directly  from  some 
other  theorem. 

15.  Theorem  1.  Adding  a  negative  number  is  equiva- 
lent to  subtracting  a  positive  number  h(iriit<J  the  same 
absolute  value.     That  is, 

a+(-b)=a-b  See  §  48,  E.  C. 

Proof.     Let  a  +  ( -  6)  =x.  (1) 

Such  ;i  number  ./•  exists  1  >v  Axiom  I. 

Adding  b  to  each  member  of  (1),  a  -\-  (—  6)  -f  b  =  x  +  b.  (2) 

By  the  associative  law  of  addition,  §  5,  and  by  §§  14,  6, 

a  +  (-  6)  +  6  =  a  +  [(-  6)  +  6]  =  a  +  0  =  a.     (3) 
From  (L?)  and  (•",)  by  §3,  x  +  b  =  a.  (i) 

From  (4),  by  the  definition  of  subtraction,  §  6, 

a  —  b  =  x.  (5) 

From  (1  )  ami  (5)  by  §  3,  a  +  (-  6)  =  a  -b. 

It  follows  from  theorem  1  that  either  of  the  symbols,  +(—  b) 
or  —  /-,  may  replace  the  other  in  any  algebraic  expression. 

16.  Corollary.  ./  -parenthesis  preceded  by  the  plus 
sign  may  be*removed  without  changing  I  If  sign  of  any 
/(■/■in  within  it.    See  *  28,  E.  C- 


THEOREMS  7 

For,  since  by  the  theorem  b  —  a  =  b  +  (  —  «),  each  subtraction  is 
reducible  to  an  addition,  so  that  the  associative  law,  §  5,  applies.     Thus 

a+  (b-c  +  d)  =  a  +  [b  +  (  - c)  +  </]  =  «  +  b  +  (- c)  +  d  =  a  +  b  -  c  +  d. 

Hence  an  expression  may  be  inclosed  in  a  parenthesis  pre- 
ceded by  the  plus  sign  without  changing  the  sign  of  any  of  its 
terms. 

17.  Theorem  2.  Subtracting  a  negative  number  is 
equivalent  to  adding  a  positive  number  having  the 
same  absolute  value.    That  is, 

a  -  (-  6)  =  a  +  b.  See  §  GO,  E.  C. 

Proof.     Let  a-(-b)  =  x.  (1) 

From  (1)  by  §2,     a  -(-  &)  +  (-&)  =  *  +  (-&).  (2) 

From  (2)  by  §§  (3  and  15,  a  =  x  +  (-b)  =  x-  b.  (3) 

Hence  by  the  definition  of  subtraction,        a  +  b—x.  (I) 

From  (1)  and  (1)  by  §  3,    a-  ( -  b)  =  a  +  b.  (5) 

It  follows  from  theorem  2  that  either  of  the  symbols  —(—6) 
or  +  6  may  replace  the  other  in  any  algebraic  expression. 

18.  Theorem  3.  A  parenthesis  preceded  by  the  minus 
sign  may  be  removed  by  changing  the  sign  of  each  term 
within  it.    That  is, 

a  _  (6  -  c  +  d)  =  a  -  b  +  c  -  d.     See  §  28,  E.  C. 

Proof     Let         a  -  (b  -  c  +  d)  =  x.  (1) 
From  (1)  by  the  definition  of  subtraction, 

a  =  x  +  (b-c  +  d).  (2) 

By  §§15,  16,  a  =  x+b  +  (-c)  +  d.  (3) 

Adding  (  —  b),  c,  and  (  —  d)  to  each  member  and  using  §  1, 

a  +  (-  b)  +  c  +(-  d)  =  x  +  b  +  (-  b)  +  c  +  (-c)  +  d  +  (-  <*)•    0±) 

From  (4),  by  §§  11,  15,  a-6  +  c-rf  =  a;.  (5) 

From  (1)  and  (5)  by  §  0,    a  - (6 -c  +  rf)  =a-i  +  c-d.  (0) 

It  follows  from  equation  (6),  read  from  right  to  left,  that  an 
expression  may  be  inclosed  in  a  parenthesis  preceded  by  a 
minus  sign,  if  the  sign  of  each  term  within  is  changed. 


8  FUNDAMENTAL   LAWS 

19.  Corollary  1.     a  -  b  =  -  (b  -  a). 

For  by  §§  15  and  1,     a  -  b  =  a  +  (-  b)  =  -  b  +  a.  (1) 

Hence  by  §  IS,  a  -b=  -(b  -a).  (2) 

20.  Corollary  2.      -a  +  (-&)  =  - (a +  6).      See  §  48,  E.  C. 

For  by  §18,  -  a  +  (-  6)=  -  [a  -  (-  ft)].  (1) 

Hence  by  §17,  -a  +  (-6)=  -  [>  +  &]•  (2) 

21.  From  the  identities 

a  +  (—  &)  =  a—b,  §  15, 

a  —  (  —  b)  =  a  +  b,  §  17, 

a-  6  =  -  (6  -  a),  §  19, 

-a  +  (-b)=-(a  +  b),  §20, 

it  follows  that  addition  and  subtraction  of  positive  and  nega- 
tive numbers  arc  reducible  to  these  operations  as  found  in  urith- 
metic,  where  all  numbers  added  and  subtracted  are  positive, 
and  where  the  subtrahend  is  never  greater  than  the  minuend. 

E.g.  5  +  (-  8)  =  5  -  8  =  -  (8  -  5)  =  -  3. 

5-(-8)=5  +  8  =  13. 

-  5-8  =  -(5  +  8)=  -  13. 

THEOREMS   ON    MULTIPLICATION   AND   DIVISION 

22.  Theorem  1.     The  product  of  any  number  and  zero  is 
Z(  vit.     That  is,  a  ■  0  =  0. 

Proof.     By  definition  oi  zero,  §  6,  a«0  =  a(6  — 6). 
By  the  distributive  law  of  multiplication,  §  10, 

a  (J>  —  b)  —  ub  —  ab, 
which  by  definition  is  zero. 

Hence  a  •  0  =  0. 

Notice  thai  by  the  commutative  law  of  multiplication,  §  8, 

a  •  0  =  0  •  a. 


THEOREMS  9 

It  follows  from  this  theorem  and  §  9,  that  a  product  is  zero 
if  any  one  of  its  factors  is  zero;  and  conversely,  by  §  11,  if  a 
product  is  zero,  then  at  least  one  of  its  factors  must  be  zero. 

23.  Corollary  1.     -  =0>  provided  a  is  not  zero. 

Since  by  the  theorem  0  =  a  •  0,  the  corollary  is  an  immediate  con- 
sequence of  the  definition  of  division  (§  11). 

24.  Corollary  2.     ^  represents  any  number  whatever. 

That  is,  ?  =  k,  for  all  values  of  k. 

Since  0  =  0  •  k,  this  is  an  immediate  consequence  of  the  definition 
of  division. 

25.  Corollary  3.  There  is  no  number  k  such  that  ^.—k, 
provided  a  is  not  zero. 

This  follows  at  once  from  k  •  0  =  0  for  all  values  of  k. 

From  §§  24,  25,.  it  follows  that  division  by  zero  is  to  be  ruled 
out  in  all  cases  unless  special  interpretation  is  given  to  the 
results  thus  obtained. 

26.  Theorem  2.  a(-b)  =  -ab.    -       See  §  63,  E.  C. 

Proof.     Let  a(-b)=x.  (1) 

By  §  2,  a(-b)+  ah  =  x  +  ab.  (2) 

By  §  10,  a[(  -  b)  +  V\  =  x  +  ab.  (3) 

By  §§  14,  22,  a.0  =  0  =  x+  ab.  (4) 

By  §14,  x  =  -ab.  (5) 

Hence,  from  (1)  and  (5)        a(—  &)=  —  ab.  ((5) 

27.  Theorem  3.  (-«)(- 6)  =06.  See  §  63,  E.  C. 

Proof.     Let  (-  a) (-  6)  =  x.  (1) 

By  §§  2  and  26,  (-a)  (-&)  +  (-  a)b  =  x-  ab.  (2) 

By  §§  10, 14,  22,    (-  a)(-  b  +  &)=0  =  x  -  ab.  (3) 

Hence,  §  14,  x  =  ab.  (4) 

From  (1)  and  (4)  (-  a)(-b)=  ab.  (5) 


10  FUNDAMENTAL   LAWS 

28.  Theorem  4.  If  the  signs  of  the  dividend  and  divisor 
are  alike,  the  quotient  is  positive;  and  if  unlike,  the  quo- 
tient is  negative.    See  §  G7,  E.  C. 

Proof.  This  theorem  is  au  immediate  consequence  of  the  definition 
of  division  and  the  identities, 

(+  a) (+  6)  =  ah.     «( -  b)  =  -  ah  and  ( -  a)  (  -  b)  =  ah. 

29.  Theorem  5.  6.a=— •  See  §  191,  E.  C. 

c  c 

Proof.  Let  x  =  -  •  a.  (1) 

By  §§  7  and  11,  ex  =  c  •   -  •  a  =  ba.  (2) 

c 

Dividing  by  c,  §  11,  x  =  — .  (3) 

c 

Hence  from  (1)  and  (3)       -  •  a  =  —  (4) 

c  c 

30.  Theorems.   *±l  =  *  +b,and°^  =  °-*-    See  §  25, 

-j7i     i-i  c  c        c  ecu 

Proof.    By  §  29,        °L+*  =  1_0i±i0  =!.(«+  6).  (1) 

c  c  c 

By  §§  10  and  29,      -   .  (a  +  b)  =  -  ■  a  +  -  .  b  =  -  +  -•  (2) 

C  CCCC 

From  (1)  and  (2),  ^^  =  «  +  !i.  (:',) 

c  c      c 

Similar]  v  we  may  show  that      = 

c         c      c 

This  theorem  states  what  is  called  the  distributive  law  of  division. 

31.  In  the  proofs  of  the  above  theorems  certain  axioms  have  been 
assumed  to  hold  in  a  more  general  form  than  the  one  in  which  they 
are  stated.  For  example,  the  commutative  law  of  addition  was  stated 
for  two  numbers  only  and  has  been  assumed  for  more  than  two. 
These  extensions  can  be  shown  to  follow  from  the  axioms  as  given. 
It  has  Likewise  been  assumed  thai  zero  and  unity,  which  were  defined 
respectively  as  a  —  a  =  0,  and  -  =  1,  are  the  same  for  all  values  of  a. 


CHAPTER   II 
FUNDAMENTAL    OPERATIONS 

32.  The  operations  of  addition,  subtraction,  multiplication, 
division,  and  rinding  powers  and  roots  are  called  algebraic 
operations. 

33.  An  algebraic  expression  is  any  combination  of  number 
symbols  (Arabic  figures  or  letters  or  both)  by  means  of  indi- 
cated algebraic  operations. 

E.g.  21,  3  -f  7,  9(b  +  c),    — — — ,  x1  +  Vy,  are  algebraic  expressions. 

34.  Any  number  symbol  upon  which  an  algebraic  operation 
is  to  be  performed  is  called  an  operand. 

All  the  algebraic  operations  have  been  used  in  the  Elementary 
Course.  They  are  now  to  be  considered  in  connection  with  the  fun- 
damental laws  developed  in  the  preceding  chapter,  and  then  applied 
to  more  complicated  expressions.  The  finding  of  powers  and  roots 
will  be  extended  to  higher  cases. 

35.  One  of  the  two  equal  factors  of  an  expression  is  called 
the  square  root  of  the  expression ;  one  of  the  three  equal  fac- 
tors is  called  its  cube  root;  one  of  the  four  equal  factors,  its 
fourth  root,  etc.  A  root  is  indicated  by  the  radical  sign  and  a 
number,  called  the  index  of  the  root,  which  is  written  within 
the  sign.     In  the  case  of  the  square  root,  the  index  is  omitted. 

E.g.  Vi  is  read  the  square  root  ofi  ;  V8  is  read  the  cube  root  of  8; 
VGi  is  read  the  fourth  root  o/64,  etc. 

36.  A  root  which  can  be  expressed  in  the  form  of  an  integer, 
or  as  the  quotient  of  two  integers,  is  said  to  be  rational,  while 
one  which  cannot  be  so  expressed  is  irrational. 


E.  g.  v8  =  2,  Va2  +  2  ah  +  IP-  =  a  +  b,  and  V|  =  §  are  rational  roots, 
while  VT  and  Va2  +  ab  +  V1  are  irrational  roots. 

11 


12  FUNDAMENTAL    OPERATIONS 

An  algebraic  expression  which  involves  a  letter  in  an  ir- 
rational root  is  said  to  be  irrational  with  respect  to  that  letter; 
otherwise  the  expression  is  rational  with  respect  to  the  letter. 

E.g.  a  +  bVc  is  rational  with  respect  to  a  and  b,  and  irrational 
with  respect  to  c. 

37.  An  expression  is  fractional  with  respect  to  a  given  letter 
if  after  reducing  its  fractions  to  their  lowest  terms  the  letter 
is  still  contained  in  a  denominator. 

E.g.    — ■ h  b  is  fractional  with  respect  to  c  and  d,  but  not  with 

c  +  d 
respect  to  a  and  b. 

38.  Order  of  Algebraic  Operations.  In  a  series  of  indicated 
operations  where  no  parentheses  or  other  symbols  of  aggrega- 
tion occur,  it  is  an  established  usage  that  the  operations  of 
finding  powers  and  roots  are  to  be  performed  first,  then  the 
operations  of  multiplication  and  division,  and  finalty  the  opera- 
tions of  addition  and  subtraction. 

E.g.  2  +  3  •  4  +  5  •  y/8  -  i2  -=-  8  =  2  +  3  •  4  +  5  •  2  -  16  -  8 

=  2  +  12  +  10-2  =  22. 

In  cases  where  it  is  necessary  to  distinguish  whether  multipli- 
cation or  division  is  to  be  performed  first,  parentheses  are  used. 

E.g.  In  6  -=-  3  x  2,  if  the  division  comes  first,  it  is  written  (G  -f-  3)  x 
2  =  4,  and  if  the  multiplication  come  first,  it  is  written  6  -=-  (3  x  2)  =  1. 

ADDITION   AND   SUBTRACTION   OF   MONOMIALS 

39.  In  accordance  with  §  10,  the  sum  (or  difference)  of  terms 
which  are  similar  with  respect  to  a  common  factor  (§  78,  E.  C.) 
is  equal  to  the  product  of  this  common  factor  and  the  sum  (or 
difference)  of  its  coefficients. 

Ex.  1.    S  ax2  +  9  ax2-  3  ax2  =  (8  +  9  -  3)  ax2  =  14  ax2. 
Ex.  2.   a\'x2  +  y2  +  bVx2  +  y2  =  (ry  +  b)  Vx2  +  y2. 


Ex 


3    x(x-\)(x-2)      r(*-l)       /x-2      -A  *(*-!) 
1-2-3  1-2  \     3  /      1 • 2 


_x+  1      ■'•(•'■  -  1)  _  (x  +  l).r(x  -  1) 
3  1-2  1.2-3 


ADDITION  AND   SUBTRACTION  13 

EXERCISES 

Perform  the  following  indicated  operations : 

1.  5.x462  —  3x*b2  —  4x-462  +  7x*b2. 

2.  3V*r  -  4  -  2Var  -  4  +  2  VaT^I  -  4  Var  -  4. 

3.  a&5c4  -  cZ65c4  +  e&V  +  /65c4. 

4.  afix4  +  5a5x4  —  5a6ari  —  Safo4. 

5.  7a?Y  +  5aY  -  9x4)f  +  5a?5?/4. 

6.  2an  +  a"-1  +  «"+1  =  an'\2a  +  1  +  a2)  =  an-J(l  +  a)2. 

7.  n(n  -  l)(n  -  2)(n  -  3)(n  -  4)  +  «(n  -  l)(n  -  2)(n  -  3). 

n(n  —  1)(«  —  2)(rc  —  3)  is  the  common  factor  and  n  —  4  and  1  are 
the  coefficients  to  be  added. 

8.  n(n  -  l)(n  -  2)(n  -  3)  (n  -  4)(w  -  5)(«  -  6) 

+  w(n  - 1)(«  -  2)(m  -  3)(n  -  4)(»  -  5). 

9.  n(n  -  l)(n  -  2)(n  -  3)  +  (n  -  1)(«  -  2)(n  -  3). 

10.  n(n  -  l)(n  -  2)  (a  -  3)(n  -  4)  +  (n  -  l)(n  -  2)(n  -  3). 

11.  (a  _  4)(6  +  3)  +  (a  -  1)(6  -  2)  +  (a  +  3) (6  +  3). 
First  add  (a  -  4)  (ft  +  3)  and  (a  +  3)  (ft  +  3). 

12.  (x  +  2y)(a>  -  2y)  +  (a?  -  Sy)(x  -  2y)  -  (2x  -y)(x-  y). 

13.  (5a  -  36) (a  -  6) (a  +  6)  +  (26  -  4a) (a  -  6)(a  +  6) 

+  (a  -  6)2(2a  -  6). 

14.  (7a?2  +  3y*)(5a>  -  y)(as  +  y)  +  (7a?2  +  3/)(*  +  y)(2  y  -  4a?) 

+  (7x2-3y2)(x  +  yy. 

15.  23-32-o  +  24.3.5. 

The  common  factor  is  23  •  3  •  5.     Hence  the  sum  is 
2«  .  3  •  5(3  +  2)  =  28  •  3  •  52. 

16.  2  •  34  •  7  +  22  •  38  •  72  -  24  •  33  •  7. 

17.  34  •  57  •  13  +  35  •  57  •  132. 


14  FUNDAMENTAL   OPERATIONS 

18.  o4  •  73  •  1 1  +  53  •  7-  •  11  -  2s  ■  3  •  5s  •  72  •  11. 

19.  3-  •  718  •  1315  +  321  •  717  •  1315  +  324 .  717  •  1315. 

20.  1-2-  3-  -n   +1  -2-3  •••  n(n  +1). 

The  dots  mean  that  the  factors  are  to  run  on  in  the  manner  indi- 
cated up  to  the  number  n.  The  common  factor  in  this  case  is 
1  •  2  •  3  •••  n,  and  the  coefficients  to  be  added  are  1  and  n  +  1.  Hence 
the  sum  is  1  •  2  •  3  •••  n(n  +  2). 

21.  1  -2  -3  ...  w   +1-2-3  ...«(n  +  l) 

+  1.2.3...n(n  +  l)(n  +  2). 

22.  1  •  2  •  3  ...  n  +  3  •  4  •  5  •  •■  n  +  5  •  6  •  7  •••  ». 

23.  n(n  -  1)  .••  (w  -  6)    +  ><(«  -  1)  ...  (n  -  6)(w  -  7). 

24.  »(w  —  1)  ...  (n  —  r)    +  n(%  —  1)  •••  (n  —  r)(n  —  r  —  1). 

25.  nanb  +  anb.  26.    w^'  ~  ^  aB_162  +  nan_162. 

0_     »(n  —  l)(w  —  2)    _o,s  .  n(w  —  1)    „_.,,, 

27.    — ^ —  — >-an  -lr-\ — i       — L  a"  -b3. 

1-2-3  1-2 

The  common  factor  is"^w  ~ — '-  an--lfi  and   the   coefficients   to  be 


added    are and  1. 


1  -2 


v(n-  l)(n  -  2)(n  -  3)        3  4      «(n  -  l)(n  -  2)        3&4 
2-3-4  2-3 


29  »fo-l)("-2)(*-3)(n-4) 

2-3.4-5 

w  n  -  1)Q,  -  2)(n  -  3) 
2-3-4 

30  n(n  -1)- (n-r  +  l)(n  -  r)         ,r+1 

2.3-r(r  +  l) 


n,  »  _  1)  ...  („  _  r  .+  1)  a„_r&r+1< 


ADDITION  AND   SUBTRACTION  15 

ADDITION   AND   SUBTRACTION   OF   POLYNOMIALS 

40.  The  addition  of  polynomials  is  illustrated  by  the  follow- 
ing example. 

Add  2a  +  3&  —  4c  and  3a  —  26  +  5  c. 

The  sum  may  be  written  thus : 

(2  a  +  3  b  -  4  c)  +  (3  a  -  2  b  +  5  c). 

By  the  associative  law,  §  5,  and  by  §  16,  we  have, 
2  a  +  3  6  —  4  c  -f  3  a  —  2  6  -f-  5  c. 

By  the  commutative  law,  §  4,  and  by  §  15,  this  becomes, 
2  a  +  3  a  +  3  6  —  2  6  —  4  c  +  5  c. 

Again  by  the  associative  law,  combining  similar  terms,  we  have, 
5  a  +  6  +  c. 

From  this  example  it  is  evident  that  several  polynomials  may 
be  added  by  combining  similar  terms  and  then  indicating  the 
sum  of  these  results. 

For  this  purpose  the  polynomials  are  conveniently  arranged  so  that 

similar  terms  shall   be  in  the   same   column.     Thus,  in   the   above 

example, 

1  2a+3  6-4c 

3a-26+5c 


5  a  +     b  + 


41.  For  subtraction  the  terms  of  the  polynomials  are  arranged 
as  for  addition.  The  subtraction  itself  is  then  performed  as  in 
the  case  of  monomials.     See  §§  17-19. 

Example.     Subtract  Ax  —  2y  +  6z  from  3x-\-6y  —  3z. 

3x  +  6  y  —  3z 
4  x  -  2  //  +  G  z 

The  steps  are : 
'3x  -4x  =  -x;    6y-(-2y)  =  8y;    -  dz  -(-f-6z)=-9«. 


16  FUNDAMENTAL    OPERATIONS 

EXERCISES 

1.  Add  8  x"  - 11  x  -  7  x2,  2  x  -  (3  x2  +  10,  -5  +  4a;3  +  9z, 
and  13  x-2  -  5  - 12  ar3. 

2.  Add    5  a3  -2a  -12  -10a2,     14-  7a  +  a2-  9a3,    3a2 

—  13  a3  +  4  —  11  a,  and  3  —  7  a  +  10  a2  +  4  a3. 

3.  From  the  sum  of  9  m3  —  3  m2  +  4  m  —  7  and  3  m2  —  4  m3 
+  2  m  +  8  subtract  4  m3  —  2  m2  —  4  +  8  m. 

4.  From  the  sum  of  x4  —  ar3  —  a2x~  —  asx  +  2  a4  and  3  ax3 
+  7  a2^2  —  5  a3x  +  2  a4  subtract  3  a.-4  +  axs  —  3  a2^2  +  a?x  —  a4. 

5.  Add  37a- 4&  — 17c  +  15d-6/-8ft  and  3c -31  a 
+  9&-5d-7i-4 ./". 

6.  Add    11  g  — 10  p  —  8 w  + 3m,  24m  —  17g  +  15p  — 13 w, 

9  n  —  G  m  —  4  q  —  7/>  —  5  n,  and  8  a  —  4^  — 12  wi  +  18  n. 

7.  From  the  sum  of  13  a  — 156  —  7c  — 11  d  and  7  a  — 66 

+  8 c  +  3 d  subtract  the  sum  of    6  d  —  56  —  7c  +  2a    and    5 c 

-  10  d  -  28  6  + 17  a. 

8.  Add  23  •  34  x3  -  25  •  32  a-2  +  22  .  33 .  7  x  +  22  .  32  •  5, 
2-  •  :;:i  Xs  -  24  •  32  •  7  x  +  24  ■  33  x2  -  2-  •  32  •  52,  and  23  •  33  cc3 
-2;-33.5  +  23.33x-24-34x2. 

9.  Add    (a  +  6  —  c)  wt  +  (a  —  6  +  c)  n,  +  (a  —  6  —  c)  A-, 

(2  a  -  3  6  +  c)  m  +  (6  -  3  a  +  c)  n  +  (4  c  +  2  6  +  a)ft, 
and  (6  - 2  c) m  +(2 a- 2c  +  6)w  +(2 6 -2a+  c)k. 

10.  From  the  sum  of  aaj8  —  bx2  +  cos  —  d    and    bx3  +  a.'c2  —  dx 

+  c  subtract  (a  —  6)  or3  +  (c  —  a)  or  —  (6  +  d)  x  —  d  +  c. 

11.  From  (m  —  n)(m  —  w)x3  +  (n  —  mfx2  —  (n  +  in) x  +  8  sub- 
tract the  sum  of  n(m—  w)^*3  —  4(w  —  m)2oj2  +  (ri  +  m)»  —  31 
and  2  (h.  —  m)2^  —  m (m  —  ri)x*  —  2  («  +  7?i)x  +  25. 

12.  Add  a"  +  2  on+1  +  a"+2  and  2  an  -  4  a"+1  +  5  an+2  and 
from  this  sum  subtract  7  a"+1  —  8  an  +  a"4"2. 


ADDITION  AND    SUBTRACTION  17 

REMOVAL    OF    PARENTHESES 

42.  By  the  theorems  of  §§  15-18,  a  parenthesis  inclosing  a 
polynomial  may  be  removed  with  or  without  the  change  of 
sign  of  each  term  included,  according  as  the  sign  —  or  +  pre- 
cedes the  parenthesis. 

In  case  an  expression  contains   signs   of  aggregation,   one 
within  another,  these  may  be  removed  one  at  a  time,  beginning 
with  the  innermost,  as  in  the  following  example : 
a  _  {h  +  c  _[<*  _  e  +f-(g  -  £)]} 
=  a  -  {b  +  c  -  Id  -  e  +  /-  g  +  K]} 
=  a-{b  +  c-d  +  e-f+g  —  7i] 
=  a  —  b  —  c  +  d  —  e+f—g+h. 

Such  involved  signs  of  aggregation  may  also  be  removed  all 
at  once,  beginning  with  the  outermost,  by  observing  the  number 
of  minus  signs  which  affect  each  term,  and  calling  the  sign  of 
any  term  +  if  this  number  is  even,  —  if  this  number  is  odd. 

Thus,  in  the  above  example,  b  and  c  are  each  affected  by  one 
minus  sign,  namely,  the  one  preceding  the  brace.  Hence  we  write, 
a  —  b  —  c. 

d  and/ are  each  affected  by  two  minus  signs,  namely  the  one  before 
the  brace  and  the  one  before  the  bracket,  while  e  is  affected  by  these 
two,  and  also  by  the  one  preceding  it.     Hence  we  write,  d  —  e  +  /• 

g  is  affected  by  the  minus  signs  before  the  bracket,  the  brace, 
and  the  parenthesis,  an  odd  number,  while  h  is  affected  by  these 
and  also  by  the  one  preceding  it,  an  even  number.  Hence  we  write 
-g  +  h. 

By  counting  in  this  manner  as  we  proceed  from  left  to  right,  we 
give  the  final  form  at  once,  a  —  b  —  c  +  d  —  e  +  /  —  g  +  h. 

EXERCISES 

In  removing  the  signs  of  aggregation  in  the  following,  either 
process  just  explained  may  be  used.  The  second  method  is 
shorter  and  should  be  easily  followed  after  a  little  practice. 

1.   7-{-4-(4-[-7])-(5-[4-5]  +  2)|. 


18  FUNDAMENTAL    OPERATIONS 

2.  _[_(7-{-4  +  9}-13)-(12-3+[-7  +  2])]. 

3.  6-(-3-[-5.+  4]  +  {7-3-(7--19)}  +  8)- 

4.  5  +  [-(-f-5-3+ll|-15)-3]+8. 


5.  4  a-  —  [3  x  —  y  —  \ 3 x  —  y  —  (x  —  y  —  x)  +  x\  —  3  ?/]. 

The  vinculum  above  y  —  x  has  the  same  effect  as  a  parenthesis,  i.e. 
-  1 1  -  x  =  -  (y  -  x). 

6.  3xi-2tf-(4:a?-\3a?-(y*-2xl) -3^-^  +  4  a2). 


7.  7  a  -  [3  a  -  [-  2  a  -  a  +  3  +  a]  -  2a-5J. 

8.  /  -  (-  2  m -n-\l-ml)-(5l-2n-[-3  m  +  »]). 

9.  2d-  [3  d  +  |2cl-(e- 5d)| -  (d  +  3 e)]. 

10.  4y-(-2y-[-3y-{-y-y^l}+2y]). 

11.  3a?-[8a;-(aj-3)-  {-2a;  +  6-8a:-l}]. 


12.  a;  -  (x  —  \  -4a;  -  [5  x  -  2  .»■  -  5]  -  [-  .r  -  a-  -3]  \). 

13.  3  »  —  {y -  [3y  +  2 2] -  (4  .r  -  [2 y  -  3  2]  -  3 y  -  2  z)  +4.r } . 

14.  a;-(-&-{-3a:-[>-2a;  +  5]-4}-[2a;-a;-3]). 

MULTIPLICATION   OF   MONOMIALS 

43.   Theorem.    The  product  of  two  powers  of  the  same 

base  is  a  power  of  that  base  whose  exjwnent  is  the  sum 
of  the  exponents  of  the  common  base.  See  §  127,  E.  C. 

Proof.    Let  b  be  any  number  and  k  and  n  any  positive  integers.    It 
is  to  be  proved  that  &* .  &»  ='&*+n 

By  the  definition  of  a  positive  integral  exponent, 
bk  =  b  •  6  -6  •••  to  k  factors, 
and  bn  —  h  •  h  •  l>  ■■■  to  n  factors. 

Hence,  6*  •  bn  =  (J>  ■  b  •••  to  k  factors) (b  ■  b  ■■•  to  n  factors) 

=  6  -6  •  &'•••  to  k  +  n  factors, 

since  the  factors  may  be  associated  in  a  single  group,  §  0. 

Hence,  by  the  definition  of  a  positive  integral  exponent,  we  have, 

6*  •  h"  =  //  • ". 


MULTIPLICATION  AND  DIVISION  19 

44.  In  finding  the  product  of  two  monomials,  the  factors  may 
be  arranged  and  associated  in  any  manner,  according  to  §§  8,  9. 

E.g.    (3 ab2)  x  (5 a2b3)  =  3  ab'2  ■  5  a2b3  §  9 

=  3  •  5  •  a  ■  a2  ■  b2 .  b3  §8 

=  (3.5)(a.a2)(&*.&8)  §9 

=  15  o365        -  by  the  theorem,  §  43 

The  factors  in  the  product  are  arranged  so  as  to  associate  those 
consisting  of  Arabic  figures  and  also  those  which  are  powers  of  the 
same  base.  This  arrangement  and  association  of  the  factors  is  equiv- 
alent to  multiplying  either  monomial  by  the  factors  of  the  other  in 
succession.     See  §  129,  E.  C. 

45.  It  is  readily  seen  that  a  product  is  negative  when  it  con- 
tains an  odd  number  of  negative  factors  ;  otherwise  it  is  positive. 

For  by  the  commutative  and  associative  laws  of  factors  the  negative 
factors  may  be  grouped  in  pairs,  each  pair  giving  a  positive  product. 
If  the  number  of  negative  factors  is  odd,  there  will  be  just  one  remain- 
ing, which  makes  the  final  product  negative. 

EXERCISES 

Find  the  products  of  the  following : 

1.  2s  •  34  •  47,  27  •  3-  •  42.  8.  a1,  a3*-»,  a?-**. 

2.  3  •  24  •  52,  5  •  22  •  5,  7  •  2s  ■  53.    9.  anbm,  cW,  al-3"b°-4m. 

3.  2  x2ys,  5  xY;  2  x*y.  10.  4  abm,  2 a3bn,  3  aGb°--m-1. 

4.  5  xy,  2  arty,  4  xy5,  x?y2.  11.  2xmym+n,  3  xm~  y»-m+2. 

5.  3a5bc,  ab2c,  a?bc4,  4ab5c.     12.  ad-'2c+2bm~3'1,  a*-*-1^"4*1. 

6.  xn,  xn-\  xn+\  2  x".  13.3  a;a+3i,  2  xa~  2hyc~3, 2  x4-2a-by2c+3. 

7.  xm+n~\  xm-"+\  x2m.  14.  a?*-sb"+1,  ax+sbv-\  3  a8&2. 
15  34a-2-26 . 2"+3_m  35-4«+26 .  om+2~n 

16.  a^+1^,  &-2*-iy**}  y"-*m         17.  7  •  23a-4,  3  •  25~2a,  5  •  23"0. 

ig      0~x-l-4y     Q   _   Ql-5x-4y     Q2  ,  92-2z 

19       32-5'"+3"  .   O-iaSb     Q2-3n+6m   _  95+3>>+5ri 

20.  (1  +  a)7-864""  •  (1  -  a)2+a"6,  (1  -  a)"-"-1  •  (1  +  a)36"0"6. 


20  FUNDAMENTAL   OPEIIATIONS 


DIVISION   OF   MONOMIALS 

46.  Theorem  1.  The  quotient  of  two  powers  of  tJie  same 
base  is  a  power  of  that  base  whose  exponent  is  the  exponent 
of  the  dividend  minus  that  of  the  divisor.    See  §  154,  E.  C. 

Proof.  Let  a  be  any  number  and  let  m  and  khe  positive  integers  of 
which  m  is  the  greater.     We  are  to  prove, 

am  ^_ak  _  am-*. 

Since  k  and  m  —  k  are  both  positive  integers,  we  have,  bv  §  43, 
a*am-*  =  at+m-*=am.  That  is,  am_*  is  the  number -which  multiplied 
by  ak  gives  a  product  am,  and  hence  by  the  definition  of  division, 


Under  the  proper  interpretation  of  negative  numbers  used 
as  exponents  this  theorem  also  holds  when  m  <  k.  This  is 
considered  in  detail  in  §  177.  We  remark  here  that  in  case 
m  =  k,  the  dividend  and  the  divisor  are  equal  and  the  quotient 
is  unity.     Hence  am  -=-  a'"  =  am~m  —  a"  —  1.     See  §  11. 

47.  Theorem  2.  In  dividing  one  algebraic  expression  bj/ 
another,  all  factors  common  to  dividend,  and  divisor  may 
be  removed  or  canceled.    See  §§  23,  156, 157,  E.  C. 

Proof.     We  are  to  show  that  —  =  -• 

bk      b 

By  definition  of  division,  §  11.   — .  bk  —  a{-,  (1) 

Also  -  -b  =  a.  (2) 

Multiplying  (2)  by  k,  -  •  bk  =  ak.  (3) 

b 

Hence  from  (1)  and  (3),  "l-  ■  bk  =  "  ■  bk.  (4) 

bk  b 

Dividing  by  bk,  «£  =  ?.  (5) 


MULTIPLICATION  AND  DIVISION  21 

-r..     .,  EXERCISES 

Divide : 

1.  4  •  24  •  37  •  52  by  3  •  23 .  34  •  5.  4.  5  a5b7cs  by  5  a4b7c4d2. 

2.  5  •  37  •  74  •  135  by  2  •  35  •  V  •  132.  5.  x2nymz"m  by  xnymzm. 

3.  3x7yh  by  2  xhjz.  6.  a8»-y»+s  by  an+6y2n+\ 

7.  ac+3d+2&d-°c+6  by  ac+2<*-4. 

8.  3a+26-7  .  53^-2«+4    by     3&+«-8  .  526-2^+3^ 

9.  ft3+2™-3»&5C7-»    by     a2+m^4„64c7-„_ 

10  /v'4a-26+l?.e-<i+6„3a+2?i+c    by     -,26— c+3aya-c+&,w3a+&-2c 

11  23a_4+76  •  33f,-4c+6  by  22a~5~7h  •  32f>-6r+7 

12.  (x  -  2)3m+1-3"  •  (cc  4-  2)-m+2-s"  by  (a;  +  2)1+2"-2"  •  (x  -  2)1-3»+2"\ 

13.  (x  —  2/)»-3^1  .  (a;  +  yy<-M+2  by  (a,  _  ^-a-sc+ra  .  ^  +  ^-3-26+7^ 

14.  (a2  -  62)3+«+"* .  (a2  -  &2)i-s*-»  by  (a2  -  62)4+*  ■  (a2  -  62)-2+26. 

MULTIPLICATION    OF  POLYNOMIALS 

48.  Theorem.  The  product  of  two  polynomials  is  equal 
to  the  sum  of  the  products  obtained  by  multiplying  each 
term  of  one  polynomial  by  every  term  of  the  other.  See 
§  86,  E.  C. 

Proof.     By  the  distributive  law,  §  10,  we  have, 

(m  +  n  +  h)(a  +  6  -f  c)=  m(a  +  b  +  c) 
+  n(a  +  b  +  c) 
+  k(a  4-  b  +  c). 
Applying  the  same  law  to  each  part,  we  have  the  product, 

ma  +  nib  +  mc  +  na  +  nh  +  nc  +  ha  +  lb  -f  fee. 
This  is  Principle  XIII  of  the  Elementary  Course. 


22  FUNDAMENTAL    OPERATIONS 

EXERCISES 

Find  the  following  indicated  products  : 

1.  (a  +  &)  (a  +  &),  i.e.  (a  +  6)2 ;  also  (a  —  b)2. 

2.  (a  +  6)  (a  +  6)  (a  +  6),  i.  e.  (a  +  &)3 ;  also  (a  -  bf. 

3.  (a  +  6)  (a  +  &)  (a  +  &)  (a  +  6),  i.  e.  (a  +  &)4 ;  also  (a  -  6)4. 

4.  (a2  +  2  ab  +  &2)  (a2  +  2  a&  +  &2)  (a  +  b). 

5.  (a2  -  2  a&  +  ft2)  (a2  -  2  a6  +  ft2)  (a  -  6). 

6.  (a2  4-  2  a&  +  &2)3 ;  also  (a  +  ft)6. 

7.  (a3  +  3  a26  +  3  aft2  +  ft3)2.       9.   (a  -  ft)(«2  +  aft  +  b2). 

8.  (a3  -  3  a2&  +  3  aft2  -  ft')2.     10.    (a  +  b)(a2  -  ab  +  ft2). 

1 1 .  (a2  +  2  aft  +  ft2)  (a4  +  4  a8&  +  6  a2b2  +  4  a&3  +  ft4) . 

12.  (a2  -  2  a&  +  ft2)  (a4  -  4  asb  +  6  a2b2  -  4  a&3  +  ft4) . 

13.  (a  —  &)(a8  +  d2b  +  aft2  +  ft3). 

14.  (a  +  &)(a3  —  a2b  +  a&2  —  6s). 

15.  (a  -  &)(a4  +  a36  +  a2&2  +  a&8  +  ft4). 

16.  (a  +  &)(a4  -  a3&  +  a2&2  -  a&s  +  ft4)- 

17.  (a  -  &)(as  +  a4ft  +  a"ft2  +  a268  +  ab4  +  &*). 

18.  (a  +  6 )( "''  -  a4ft  +  a"ft2  -  a2lr  +  ,/ft4  -  ft5). 

19.  (1  -  r)(a  +  ar  +  ar2  +  ar3). 

20.  (1  —  r)(a  +  ca-  +  or  +  ar8  +  mA  +  a?-5). 

21.  (a  +  &  +  c)2.         22.    (a  +  &  -  c)2.         23.    (a  -  6  -  c)2. 

24.  From  Exs.  21-23  deduce  a  rule  for  squaring  a  trinomial. 

25.  (a;  +  >/  +z  +  v)2.  26.     (aj  —y  +  z—  r)2. 

27.  From  Exs.  25,  26  deduce  a  rule  for  squaring  a  polynomial. 

28.  (a  +  &  +  c)(a  4-  6  -  c)(a  -  ft  +  c)(&  -  a  +  c). 

29.  (ab  +  ac  +  bc)(ab  +  ac  —  &c)(a&  —  ac  +  6c)i  ac  +  be  —  ab ) . 

30.  (a-  b  +  c  +  d)(a  4-  6  +  c  —  ri)(a  4-  6  —  c  4-  d) 

(-a  +  6  +  c  +  d). 


MULTIPLICATION  AND  DIVISION  23 

31.  (4 x3  -  6 xy  +  9  y2)(2  x  +  3  ?/)(4 «2  +  6 a;#  -f  9 y2)(2 x-oy). 

32.  Collect  in  a  table  the  following  products : 

(a+b)2,         (a -hf,         (a  +  b)3,         (a  -  bf,         (a  +  b)\ 
(a  -  b)\         (a  +  b)5,         (a  -  by,         (a  +  b)G,         (a  -  b)6. 

33.  From  the  above  table  answer  the  following  questions: 
(«)  How  many  terms  in  each  product,  compared  with  the 

exponent  of  the  binomial  ? 

(l>)  Tell  how  the  signs  occur  in  the  various  cases. 

(c)  How  do  the  exponents  of  a  proceed  ?  of  b  ? 

(d)  Make  a  table  of  the  coefficients  alone  and  memorize  this. 
E.g.    For  (a  +  by,  they  are  1,  5,  10,  10,  5,  1. 

34.  Make  use  of  the  rules  in  Exs.  24,  27,  33  to  write  the  fol- 
lowing products  :  (a)  (2  x  —  3  y  +  4  z)2,  (b)  (}m2-|nJ-3r)2, 

(c)   (4  a.»  —  2  ay  +  3  m  —  nf. 

^  (3«-62)4-  V°     ^  (*>(9*-2y)«. 

(/)  (i  «»  -  1  W-        (0  (  o  3  _  yV  (0  (1-2  x)\ 

(g)  (2  m  -  n)5.  V  LV  (m)  (jL+3  xym 

DIVISION    OF  POLYNOMIALS 

49.  According  to  the  distributive  law  of  division,  §  30,  a 
polynomial  is  divided  by  a  monomial  by  dividing  each  term  sepa- 
rately by  the  monomial.     See  §  25,  E.  C. 

r,        ab  +  ac  —  ad      ab  ,  ac      ad      ,    ,  7 

E.g.    -  = 1 -  =  b  +  c  —  d. 

a  a        a        a 

A  polynomial  is  divided  by  a  polynomial  by  separating  the 
dividend  into  polynomials,  each  of  which  is  the  product  of  the 
divisor  and  a  monomial.  Each  of  these  monomial  factors  is  a 
part  of  the  quotient,  their  sum  constituting  the  whole  quotient. 
The  parts  of  the  dividend  are  found  one  by  one  as  the  work  pro- 
ceeds.   See  §§  161-163,  E.C.    This  is  best  shown  by  an  example. 


24  FUNDAMENTAL   OPERATIONS 

Dividend,  o4  +     «3  —  4a2  +  5  a  —  3  la2  +  2  a  —  3,  Divisor. 


-  a3  - 

-a2 

+  5a  - 

-3 

-  a3  - 

-2  a2 

+  3  a 

a2 

+  2a- 

-  3 

a2 

+  2a- 

-3 

1  st  part  of  dividend  :   «4  +  2  a3  —  3  a2  ]a2  — a  +  1,      Quotient. 

2d  part  of  dividend  : 

3d  part  of  dividend  : 

0 

The  three  parts  of  the  dividend  are  the  products  of  the 
divisor  and  the  three  terms  of  the  quotient.  If  after  the  suc- 
cessive subtraction  of  these  parts  of  the  dividend  the  remain- 
der is  zero,  the  division  is  exact.  In  case  the  division  is  not 
exact,  there  is  a  final  remainder  such  that 

Dividend  =  Quotient  x  divisor  -f-  Remainder. 

In  symbols  we  have  D  =  Q  •  d  -f  R. 

t^.     .j  EXERCISES 

Divide  : 

1.  x'  +  5  xAy  -f  10  arty2  +  1 0  xhf  +  5  xy*  +  f  by  .r2  +2  xy+y2. 

2.  .r*-j-.,-y  +  /  by  a4  —  x2y2-\-y\    6.  as8  —  ^  by  .r  —  y\ 

3.  Xs  — y5  by  x  —  y.  7.  a8  -f  b6  by  a2  +  62. 

4.  a8— j/8  by  37*+  tfy+xy^+y*.     8.  ,v'"-)/9byf°-f. 

5.  .r'  +  .v11  by  x*-xy-\-y2.  9.  a10— afl&5+&10  by  a2- a&+&2. 

10.  2  a4  -  3  .rV;  +  6  .r/r  -  a*8  +  6  64  by  a-2  -  2  a?6  +  3  b2. 

11.  2  .r'''  -  5  .i-5  +  ( ;  .e4  -  6  x8 + ( i  .*'-  -  4  x  + 1  by  a4  -  .i-3 + a-2  -  x + 1 . 

12.  26  a868  +  a6  +  6  b6  -  5  a'6  -  17  ab5  -  2  cW  -  a?b* 

by  a2-3&2-2a&. 

13.  a4  +  2.r3-7.r-8a  +  12  by  a-2 -3 a +  2. 

14.  4  //-■  +  4  a6  4.  «-•  _  12  6c  -  6 ac  +  9 c2  by  2b  +  a-Sc. 

15.  .i'4  -)-  4  x> f  —  4  :cyz  +  3  y*  +  2  ?/2z  —  z2  by  a-2  —  2  a-//  +  3  //-'— z. 

16.  «262c  +  3  crb3  -  3  «&<-"■  -  aV  +  //'  -  4  &V  +  3  a63c 

+  3  6c4—  3  a?b&  by  b'2  —  c2. 


CHAPTER   III 

INTEGRAL   EQUATIONS  OF  THE   FIRST    DEGREE  IN  ONE 
UNKNOWN 

50.  When  in  an  algebraic  expression  a  letter  is  replaced  by 
another  number  symbol,  this  is  called  a  substitution  on  that  letter. 

E.g.  In  the  expression,  2  a  +  5,  if  a  is  replaced  by  3,  giving  2  •  3  +  5, 
this  is  a  substitution  on  the  letter  a. 

51.  An  equality  containing  a  single  letter  is  said  to  be 
satisfied  by  any  substitution  on  that  letter  which  reduces  both 
members  of  the  equality  to  the  same  number. 

E.g.     4  x  +  8  =  24  is  satisfied  by  x  =  4,  since  4  •  4  +  8  =  24. 

We  notice,  however,  that  the  substitution  must  not  reduce  the 
denominator  of  any  fraction  to  zero. 

x2  —  4 
Thus  x  =  2  does  not  satisfy  —     —  =  8  although  it  reduces  the  left 

0 
member  of  the  equation  to  -,  which  by  §  21  equals  8  or  any  other 

number  whatever. 

On  the  other  hand,  x  =  6  satisfies  this  equation,  since 

62-4      32 

o~T  =  4  =  8' 

52.  An  equality  in  two  or  more  letters  is  satisfied  by  any 
simultaneous  substitutions  on  these  letters  which  reduce  both 
members  to  the  same  number. 

E.g.  6  a  +  3  b  =  15  is  satisfied  by  a  =  2,  6  =  1 ;  a  =  -,  6  =  2 ;  a  =  1, 
6  =  3,  etc. 

,2  1. 

-  is  satisfied  by  x  =  3,  y  —  1,  but  is  not  satisfied  by 


x*  —  yA 


x-  +  2  xy  +  y- 

any  values  of   x   and  y  such  that  x  =  —  y,   since   these   reduce   the 
denominator  (and  also  the  numerator)  to  zero.     See  §  24. 

25 


26      INTEGRAL  EQUATIONS   OF  THE  FIRST  DEGREE 

53.  An  equality  is  said  to  be  an  identity  in  all  its  letters,  or 
simply  an  identity,  if  it  is  satisfied  by  every  possible  substitur 
tion  on  these  letters,  not  counting  those  which  make  any 
denominator  zero. 

If  an  equality  is  an  identity,  both  members  will  be  reduced 
oO  the  same  expression  when  all  indicated  operations  are  per- 
formed as  far  as  possible. 

The  members  of  an  identity  are  called  identical  expressions. 

Thus  in  the  identity  (a  +  6)2=a2+2  ab  -f  //-,  performing  the  indi- 
cated operation  in  the  first  member  reduces  it  to  the  same  form  as  the 
second. 

54.  An  equality  which  is  not  an  identity  is  called  an  equa- 
tion of  condition  or  simply  an  equation. 

The  members  of  an  equation  cannot  be  reduced  to  the  same 
expression  by  performing  the  indicated  operations. 

E.g.  (x  —  2)(x  —  3)  =  0  cannot  be  so  reduced.  This  is  an  equation 
winch  is  satisfied  by  x  =  2  and  x  —  3.     See  §  22. 

55.  In  an  equation  containing  several  letters  any  one  or  more 
of  them  may  be  regarded  as  unknown,  the  remaining  ones  being 
considered  known.  Such  an  equation  is  said  to  be  satisfied  by 
any  substitution  on  the  unknown  letters  which  reduces  it  to  an 
identity  in  the  remaining  letters. 

E.g.  x2  —  t'2  =  sx  +  st  is  an  equation  in  s,  x,  or  t,  or  in  any  pair  of 
these  letters,  or  in  all  three  of  them. 

As  an  equation  in  x  it  is  satisfied  by  x  =  s  +  /,  since  this  substitu- 
tion reduces  it  to  the  identity  in  s  and  t, 

s2  +  2  st  =  s2  +  2  st. 

As  an  equation  in  s  it  is  satisfied  by  s  =  x  —  t,  since  this  substitu- 
tion reduces  it  to  the  identity  in  x  and  /. 

X2_   t*  =  X*-   ,2. 

Any  number  expression  which  satisfies  an  equation  in  one 
unknown  is  called  a  root  of  the  equation. 

E.g.  s  +  t  is  a  root  of  the  equal  ion  ./•-  —  t-  =  sx  +  st,  when  x  is  the 
unknown,  and  x  —  t  is  a  root  when  .s-  is  the  unknown. 


EQUIVALENT  EQUATIONS  27 

56.  An  equation  is  rational  in  a  given  letter  if  every  term  in 
the  equation  is  rational  with  respect  to  that  letter. 

An  equation  is  integral  in  a  given  letter  if  every  term  is 
rational  and.  integral  in  that  letter. 

57.  The  degree  of  a  rational,  integral  equation  in  a  given 
letter  is  the  highest  exponent  of  that  letter  in  the  equation. 

In  determining  the  degree  of  an  equation  according  to  this  defini- 
tion it  is  necessary  that  all  indicated  multiplications  be  performed  as 
far  as  possible. 

E.g.  (x  —  2)  (x  —  3)  =  0  is  of  the  2d  degree  in  x,  since  it  reduces  to 
x'2  —  5  x  +  6  =  0. 

EQUIVALENT  EQUATIONS 

58.  Two  equations  are  said  to  be  equivalent  if  every  root  of 
either  is  also  a  root  of  the  other. 

59.  Theorem  1.  If  one  rational,  integral  equation  is 
derived  from  another  by  performing  the  indicated  opera- 
tions, then  the  two  equations  are  equivalent.  See  §  36, 
E.  C. 

Proof.  In  performing  the  indicated  operations,  each  expression  is 
replaced  by  another  identically  equal  to  it.  Hence  any  expression 
which  satisfies  the  given  equation  must  satisfy  the  other  and 
conversely. 

E.g.  10  x  =  50  is  equivalent  to  3  x  +  7  x  =  50,  since  3  x+  7  x  =  10  x ; 
and  8(2  x  —  3  y)  =  2  y  —  1  is  equivalent  to  16  x  —  2±y  =  2  y  —  1,  since 
8(2 x  -  3t/)ee  16 x  -24y. 

60.  Theorem  2.  If  any  equation  is  derived  from  another 
by  adding  the  same  expression  to  each  member,  or  by  sub- 
tracting the  same  expression  from  each  member,  then  the 
equations  are  equivalent.     See  §  36,  E.  C. 

Proof.  For  simplicity  of  statement  we  prove  the  theorem  for  the 
case  where  the  original  equation  contains  only  one  unknown,  the 
proof  in  the  other  cases  being  similar. 

Let  M=N  (1) 

be  an  equation  involving  one  unknown,  x,  and  let  A  be  an  expression 
which  may  or  may  not  involve  x. 


28      INTEGRAL    EQUATIONS   OF  THE  FIRST  DEGREE 

We  are  to  show  that  equation  (1)  is  equivalent  to 

M  +  A=N+\.  (2) 

and  also  to  M  -  A  =  N  —  A.  (3) 

(a)  Let  rj  be  a  root  of  (1).  Then  substituting  xy  for  x  in  (1) 
makes  M  and  N  identical.  Since  A  =  A  for  any  value  of  x  it  follows, 
§  "_'.  that  the  substitution  of  x1  reduces  M  +  A  and  A*  +  A  to  identical 
expressions.  That  is,  x  =  xx  satisfies  equation  (2).  Hence  any  root 
of  (1)  is  also  a  root  of  (2). 

(h)  Again  if  x1  is  a  root  of  (2),  its  substitution  reduces  M  +  A  and 
N  +  A  to  identical  expressions,  and  hence  by  §  6,  it  also  reduces 
M  +  A  —  A  and  N  +  A  —  A  to  identical  expressions.  That  is,  x  =  xx 
satisfies  equation  (1).  Hence  any  root  of  (2)  is  also  a  root  of  (1). 
From  (a)  and  (b)  it  follows  that  equations  (1)  and  (2)  are  equivalent. 

In  like  manner  (1)  and  (3)  are  shown  to  be  equivalent. 

61.  Corollary.  Any  equation  can  be  reduced  to  an 
equivalent  equation  of  the  form  R  —  0. 

For  if  an  equation  is  in  the  form  M  —  N,  then  by  theorem  1  it 
is  equivalent  to  M  -  i\r  =  N  —  N  =  0,  which  is  in  the  form  R  =  0. 

62.  Theorem  3.  i/1  one  equation  is  derived  from  <tu<>llicr 
by  multiplying  or  dividing  each  member  by  the  same  ex- 
pression, then  the  equations  are  equivalent,  provided  the 
original  equation  is  not  multiplied  or  divided  by  zero  or 
by  an  expression  containing  the  unknown  of  theequation. 

See  §  36,  E.  C,  and  the  note  following  it. 

Proof.  Again  consider  the  case  where  the  original  equation  con- 
tains only  one  unknown. 

Let  A  be  an  expression  not  containing  x,  and  different  from  zero. 
AYe  are  to  show  that  M  =  N  (1) 

is  equivalent  to  M  ■  A  =  N  ■  A,  (2) 

and  also  to  S± -±L.  (3) 

A       A  W 

If  r,  is  a  root  of  (1),  its  substitution  makes  .1/"  and  X  identical, 
and  hence  also  M  .  A  and  .V  •  .1  by  §  7.  That  is.  r,  is  a  root  of  (2). 
Similarly,  if  .;-,  is  a  root  of  (2).  then,  by  §  11,  it  is  a  root  of  (1). 

Hence  (1)  and  (2)  are  equivalent.  In  like  manner,  we  may  show 
(1)  and  (3)  equivalent. 


EQUIVALENT  EQUATIONS  29 

63.  The  ordinary  processes  of  solving  equations  depend 
upon  theorems  1,  2,  and  3,  as  is  illustrated  by  the  following 
examples : 

Ex.  1.  (x  +  4)  (./;  +  5)  =  (x  +  2)  (x  +  6).  (1) 

x1  +  9  x  +  20  =  x2  +  8  x  +  12.  (2) 

x  =  -  8.  (3) 

By  theorem  1,  (1)  and  (2)  are  equivalent,  and  by  theorem  2,  (2) 
and  (3)  are  equivalent.     Hence  (1)  and  (3)  are  equivalent.      That  is, 

—  8  is  the  root  of  (1). 

Ex.2.  |X  +  |  =  4.  (1) 

2  x  +  4  =  12.  (2) 

2  x  =  8.  (3) 

x  =  4.  (4) 

By  theorem  3,  (1)  and  (2)  are  equivalent.  By  theorem  2,  (2)  and 
(3)  are  equivalent.  By  theorem  3,  (3)  and  (4)  are  equivalent. 
Hence  (1)  and  (4)  are  equivalent  and  4  is  the  solution  of  (1). 

These  theorems  are  stated  for  equations,  but  they  apply 
equally  well  to  identities,  inasmuch  as  the  identities  are 
changed  into  other  identities  by  these  operations. 

64.  If  an  identity  is  reduced  to  the  form  R  —  0,  §  61,  and  all 
the  indicated  operations  are  performed,  then  it  becomes  0  =  0. 
See  §  53.  Conversely,  if  an  equality  may  be  reduced  to  the 
form  0  =  0,  it  is  an  identity.  This,  therefore,  is  a  test  as  to 
whether  an  equality  is  an  identity. 

E.g.  (x  +  4)'2  =  x2  +  8x  +  16  is  an  identity,  since  in  x2  +  8x  +  16 

—  x2  —  8x  —  16  =  0  all  terms  cancel,  leaving  0  =  0. 

EXERCISES 

In  the  following,  determine  which  numbers  or  sets  of  num- 
bers, if  any,  of  those  written  to  the  right,  satisfy  the  corre- 
sponding equation. 

Remember  that  no  substitution  is  legitimate  which  reduces 
any  denominator  to  zero. 


30     INTEGRAL   EQUATIONS   OF  THE  FIRST  DEGREE 

1.  4(a»-l)(a>-2)(a;-3)=3(a;-2)(a>-3).  1,2,3,4. 

2.  t^>=(x-4:)(x  +  6).  2,4,6. 

x  +  5 

X  +  3  3  r>     o      ! 

3.  — ^        =  a;  —  -  •  2,  d,  i. 

Var'  +  7  A 

4.  (^-3)^-2)  =  aro_5a;  +  6,  2,3,0,-2. 
x2  —7x  +  10 

5.  «2  +  9a  +  20      (a+4)(a_4)(a+5)j        4>_4)5)_5. 
a-  +  8  a  + 16 

fa=0,    (a  =  4,    |«  =  2, 
'•»•  +  «  =  *  |6«S.    U  =  0.    |&=2. 

7    369(a-6)_fl  |  b  Ja  =  °J    J"  =  1>    f«  =  5> 


aa  +  62  16  =  0.    [6=1.    [6  =  4. 

s=l,    f.r=l,    lx  =  2, 

3  ftt  =  l,       ftt  =  l,       ftt  =  — 1, 


8.   ^^^  =  (x-2)(i/-l). 

K-2/  1^  =  0.    U  =  l-    U 


?r  —  v 


9. -  =  (u*  +  uv  +  v*)(u  —  v). 

v?  —  v2  I  '" 


-1.    [v  =  D.    \v  =  0. 


1Q    (r-s)(r  +  s)(r2  +  r)  =  (/.2  _  ^  f _  3  r  =  lf  ,  =  t . 

?"3  +  rs2  —  rs  —  s3 

7-  =  1,  s  =  —  1 ;     r=  2,  s  =  —  2 ;     r  =  a,  s  =  —  a. 

11.  a  +  6  +  c  =  6.     a  =  l,6=2,c  =  3; 

a  =  3,  6  =  3,  c  =  0  ;     a  =  10,  6  =  0,  c  =  -  4. 

a  —  b  +  c         3a  —  2c  +  56  +  2  ,,,      n^     a 

12.  .  T=  =  ~    T7T~     —'     a  =  8,  6  =  0,  c  =  6; 

a  =  l,  6  =  4,  c=2;     a  =  0,  6  =0,  c=-4. 

13    (a-6)(6-c)(c-a)  =  (6_6(c_a)   a==2>6  =  1>c  =  1, 

ac  —  6c  —  a-  +  6a 

a  =  3,  6  =  2,  c  =  3 ;     a  =  6,  6  =  6,  c  =  0. 


EQUIVALENT  EQUATIONS  31 

14.    (as  -  2!)  (a;  -  y)  (y  -  z)  =  8  xyz(x2  -  if)  (y2  -  z2)  (z2  -  x2). 
x=  1,  y  =  1,  z  —  1 ;     ar  =  l,  y  =  0,  z  =  l;     a;  =  1,  ?/  =  2,  z  =  3. 


15.  x34-3x2?/4-3^24-?/3  =  (.c4-?/)3.  y  0    • 

[y  =  l.    [y  =  2.   I  ?/ =  4. 

16.  Show  by  reducing  the  equality  in  Ex.  15  to  the  form 
R=  0  that  it  is  satisfied  by  any  pair  of  values  whatsoever  for 
x  and  y,  e.g.,  for  x  =  348764,  y  =  594021.  What  kind  of  an 
equality  is  this  ? 

Which  of  the  following  four  equalities  are  identities? 

17.  12{x  +  y)2  +  17(x  +  y)-7  =  (3x  +  3y-l)(4,x  +  4,y  +  7). 

1 8.  ^IzlA5  _  tt4  +  as6  +  a262  +  abs  +  &4< 

a  —  b 

19.  ?£±&!  =  a4  _  a3b  +  a2&2  _  ab3  +  &4 

20.  2  (a  -  6)2  +  5(a  +  &)  +  8  ab  =  (2  a  4-  2  &  4-  l)(a  4-  b  +  1). 

Solve  the  following  equations,  and  verify  the  results : 

21.  (2  a  +  3)(3  a  -  2)  =  a2  4-  a  (5  a  +  3). 

22.  6  (6  -  4)2  =  -  5  -  (3  -  2  b)2  -  5  (2  +  &)  (7-2  6). 

23.  (y  -  3)2  +  (.V  -  4)2  -  0/  -  2)2  -  (i,  -  3)2  =  0. 

24.  (as  -  3)  (3  x  +  4)  -  (as  -  4)  (x  -  2)  =  (2a:  + 1)  (x-  6). 

25.  2  (3  r  -  2)  (4  r  4-  1 )  +  (r  -  4)3  =  (r  +  4)3  -  2. 

26.  a3  —  c  +  bsc  +  abc  =  b.     (Solve  for  c.) 

27.  (&_2)2(&-t/)-3%4-(2^4-1)(&-1)  =  3-2&.      (Findy.) 

28.  2  (12  -  a;)  4-  3(5  x  -  4)  +  2  (16  -  a:)  =  12(3  4-  x). 

29.  (b-a)x-(a  +  b)x+4:a2  =  0.  (Find  x.) 

30.  (a;— a)(&— c)4-(&— a)(a>-c)  — (a— c)(a?— &)=0.    (Find*.) 

31.  r3v  +  s3v  —  3  r  —  3  s  4-  3  v  (rs  +  rs2)  =  0.  (Find  i>.) 


32      INTEGRAL    EQUATIONS    OF  THE  FIRST   DEGREE 

32.  (aj  -  3)  (x  -  7)  -  (a?  -  5)  (a;  -  2)  + 12  =  2  (a?  - 1). 

33 .  (a  +  hf  +  (a-  -  b)  (x  -a)-  (x  +  a)  (x  +  6)  =  0.      ( Find  x.) 

34.  ny  (2/  +  »)  -  (y  +  m)  (//  +  »)  (w  +  »)  +  ™?/  (2/  +  m)  =  0- 

(Find  //.) 

35.  (n  +  i)  (J  -  i  +  k)  -  (n  -  i)  (i  -j  +  k)  =  0.  (Find  ».) 

36.  ^(5a:-l)+fV(2-3.r)  +  i(4  +  a-)  =  |(l+2.r)-TV 

37.  (J  — m)(»— n)  +  22(i»  +  n)  =  (J  +  m)(*+n).        (Findz.) 

38.  a(x  —  b)  —  (a  +  &)(aj  +  b  —  a)=b(x—a)+a2—b2.    (Find  x.) 

39.  (m  +  n)(n  +  &  —  y)  +  (n  —  m)(&— y)=n(m+b).    (Find  y.) 

..     3(2a-3  6)      2(3a-5&)  ,  5(a-b)      b  ._.    ,     . 

40.  -^-         --^-g-     2  +  -L_J  =  _.  (Fmda.) 

Solve  each  of  the  following  equations  for  each  letter  in  terms  of 
the  others. 

41.  l(W+w')  =  l'W'.  43.   m,.%(t.2-t)  =  (m  +  ml)(t  —  t1). 

42.  (v —  n)d=(v—n])d1.      44.   (m  +  m^Qi  — t) —  lm2+m.,t. 

PROBLEMS 

1.  What  number  must  be  added  to  each  of  the  numbers  2, 
26,  10  in  order  that  the  product  of  the  first  two  sums  may 
equal  the  square  of  the  last  sum  ? 

2.  What  number  must  be  subtracted  from  each  of  the  num- 
bers 9,  12,  18  in  order  that  the  product  of  the  first  two  re- 
mainders may  equal  the  square  of  the  last  remainder  ? 

3.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c  in  order  that  the  product  of  the  first  two  sums  may 
equal  the  square  of  the  last  ? 

Note  that  problem  1  is  ;t  special  case  of  3.  Explain  how  2  may 
also  be  made  a  special  case  oE  •'). 

4.  What  number  musl  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  product  of  the  first  two  sums  may 
equal  the  product  of  the  last  two  '.' 


PROBLEMS    IN    ONE    UNKNOWN  33 

5.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  4. 

6.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  sum  of  the  squares  of  the  first  two 
sums  may  equal  the  sum  of  the  squares  of  the  last  two  ? 

7.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  6. 

8.  What  number  must  be  added  to  each  of  the  numbers 
a,  b,  c,  d  in  order  that  the  sum  of  the  squares  of  the  first  two 
sums  may  be  k  more  than  twice  the  product  of  the  last  two  ? 

9.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  8. 

10.  The  radius  of  a  circle  is  increased  by  3  feet,  thereby  in- 
creasing the  area  of  the  circle  by  50  square  feet.  Find  the 
radius  of  the  original  circle. 

The  area  of  a  circle  is  irr2.     Use  3}  for  it. 

11.  The  radius  of  a  circle  is  decreased  by  2  feet,  thereby 
decreasing  the  area  by  36  square  feet.  Find  the  radius  of  the 
original  circle. 

12.  State  and  solve  a  general  problem  of  which  10  is  a 
special  case. 

13.  State  and  solve  a  general  problem  of  which  11  is  a 
special  case. 

How  may  the  problem  stated  under  12  be  interpreted  so  as  to 
include  the  one  given  under  13? 

14.  Each  side  of  a  square  is  increased  by  a  feet,  thereby 
increasing  its  area  by  b  square  feet.  Find  the  side  of  the 
original  square. 

Interpret  this  problem  if  a  and  b  are  both  negative  numbers. 

15.  State  and  solve  a  problem  which  is  a  special  case  of  14, 
(1)  when  a  and  b  are  both  positive,  (2)  when  a  and  b  are  both 
negative. 


34      INTEGRAL    EQUATIONS    OF   THE  FIRST   DEGREE 

16.  Two  opposite  sides  of  a  square  are  each  increased  by  a 
feet  and  the  other  two  by  b  feet,  thereby  producing  a  rectangle 
whose  area  is  c  square  feet  greater  than  that  of  the  square. 
Find  the  side  of  the  square. 

Interpret  this  problem  when  a,  b,  and  c  are  all  negative  numbers. 

17.  State  and  solve  a  problem  which  is  a  special  case  of  16, 
(1)  when  a,  b,  and  c  are  all  positive,  (2)  when  a,  b,  and  c  are 
all  negative. 

18.  A  messenger  starts  for  a  distant  point  at  4  a.m.,  going 
5  miles  per  hour.  Four  hours  later  another  starts  from  the 
same  place,  going  in  the  same  direction  at  the  rate  of  9  miles 
per  hour.  When  will  they  be  together  ?  When  will  they  be 
8  miles  apart  ?     How  far  apart  will  they  be  at  2  p.m.  ? 

For  a  general  explanation  of  problems  on  motion,  see  p.  115,  E.  C. 

19.  One  object  moves  with  a  velocity  of  i\  feet  per  second 
and  another  along  the  same  path  in  the  same  direction  with  a 
velocity  of  v2  feet.  If  they  start  together,  how  long  will  it  re- 
quire the  latter  to  gain  n  feet  on  the  former  ? 

From  formula  (2),  p.  117,  E.  C,  we  have   t  =  — - — 

Discussion.  Ifw2>i'j  and  n  >  0,  the  value  of  t  is  positive,  i.e.  the 
objects  will  be  in  the  required  position  some  time  after  the  time  of 
star!  ing. 

If  Vz  <  '"l  and  n  >  0,  the  value  of  (  is  negative,  which  may  be  taken 
to  mean  that  if  the  objects  had  been  moving  before  the  instant  taken 
in  tlic  problem  as  the  time  of  starting,  then  they  would  have  been  in 
the  required  position  some  time  earlier. 

Jiv2  =  vi  and  n=£0,  the  solution  is  impossible.  See  §  25.  This 
means  that  the  objects  will  never  be  in  the  required  position.  If 
vt  ■=  i>2  and  n  =  0,  the  solution  is  indeterminate.  See  §24.  This  may 
be  interpreted  to  mean  that  the  objects  are  always  in  the  required 
position. 

20.  State  and  solve  a  problem  which  is  a  special  case  of  19 
under  each  of  the  conditions  mentioned  in  the  discussion. 

21.  At  what  time  after  5  o'clock  are  the  hands  of  a  clock 
first  in  a  straight  line  ? 


PROBLEMS    IN    ONE    UNKNOWN  35 

22.  Saturn  completes  its  journey  about  the  sun  in  29  years 
and  Uranus  in  84  years.  How  many  years  elapse  from  con- 
junction to  conjunction?     See  figure,  p.  119,  E.  C. 

23.  An  object  moves  in  a  fixed  path  at  the  rate  of  vr  feet 
per  second,  and  another  which  starts  a  seconds  later  moves  in 
the  same  path  at  the  rate  of  v3  feet  per  second.  In  how  many 
seconds  will  the  latter  overtake  the  former  ? 

24.  In  problem  23  how  long  before  they  will  be  d  feet 
apart  ? 

If  in  problem  24  d  is  zero,  this  problem  is  the  same  as  23.  If  d  is 
not  zero  and  a  is  zero,  it  is  the  same  as  problem  19. 

25.  A  beam  carries  3  weights,  one  at  each  end  weighing 
100  and  120  pounds  respectively,  and  the  third  weighing  150 
pounds  2  feet  from  its  center,  where  the  fulcrum  is.  What  is 
the  length  of  the  beam  if  this  arrangement  makes  it  balance  ? 

For  a  general  explanation  of  problems  involving  the  lever,  see  pp. 
120-122,  E.  C. 

26.  A  beam  whose  fulcrum  is  at  its  center  is  made  to  bal- 
ance when  weights  of  GO  and  80  pounds  are  placed  at  one  end 
and  2  feet  from  that  end  respectively,  and  weights  of  50  and 
100  pounds  are  placed  at  the  other  end  and  3  feet  from  it 
respectively.     Find  the  length  of  the  beam.    . 

27.  How  many  cubic  centimeters  of  matter,  density  4.20, 
must  be  added  to  150  ccm.  of  density  8.10  so  that  the  density 
of  the  compound  shall  be  5.4?     See  §  99,  E.  C. 

28.  How  many  cubic  centimeters  of  nitrogen,  density 
0.001255,  must  be  mixed  with  210  ccm.  of  oxygen,  density 
0.00143,  to  form  air  whose  density  is  0.001292? 

29.  A  man  can  do  a  piece  of  work  in  16  days,  another  in  18 
days,  and  a  third  in  15  days.  How  many  days  will  it  require 
all  to  do  it  when  working  together? 

30.  A  can  do  a  piece  of  work  in  a  days,  B  can  do  it  in  b 
days,  C  in  c  days,  and  I)  in  d  days.  How  long  will  it  require 
all  to  do  it  when  working  together? 


CHAPTER   IV 

INTEGRAL  LINEAR   EQUATIONS   IN  TWO   OR   MORE 
VARIABLES 

INDETERMINATE    EQUATIONS 

65.  If  a  single  equation  contains  two  unknowns,  an  unlimited 
number  of  pairs  of  numbers  may  be  found  which  satisfy  the 
equation. 

E.g.  In  the  equation,  y  =  2  x  +  1,  by  assigning  any  value  to  x,  a 
corresponding  value  of  y  may  be  found  such  that  the  two  together 
satisfy  the  equation. 

Thus,  x  =  —  3,  y  =  —  5 ;  x  =  0,  y  =  1  ;  x  =  2,  y  =  5,  are  pairs  of 
numbers  which  satisfy  this  equation. 

Fortius  reason  a  single  equation  in  two  unknowns  is  called 
an  indeterminate  equation,  and  the  unknowns  are  called  varia- 
bles. A  solution  of  such  an  equation  is  any  pair  of  numbers 
which  satisfy  it. 

A  picture  or  map  of  the  real  (see  §§  135,  195)  solutions  of 
an  indeterminate  equation  in  two  variables  may  be  made  by 
means  of  the  graph  as  explained  in  §§  107,  108,  E.  C. 

66.  The  degree  of  an  equation  in  two  or  more  letters  is  the 
sum  of  the  exponents  of  those  letters  in  that  one  of  its  terms 
in  which  this  sum  is  greatest.     See  §  110,  E.  C. 

E.g.  y  =  2  x  +  1  is  of  the  first  degree  in  x  and  y.  y-  =  2  x  +  y  and 
y  =  2  xy  +  3  are  each  of  the  second  degree  in  x  and  y. 

An  equation  of  the  first  degree  in  two  variables  is  called  a 
linear  equation,  since  it  can  be  shown  that  the  graph  of  every 
such  equation  is  a  straight  line. 

36 


INDETERMINATE  EQUATIONS  37 

67.  It  is  often  important  to  determine  those  solutions  of  an 
indeterminate  equation  which  are  positive  integers,  and  for  this 
purpose  the  graph  is  especially  useful. 

Ex.  1.     Find  the  positive  integral  solutions  of  the  equation 

3  x  +  7  y  =  42. 

Solution.  Graph  the  equation  carefully  on  cross-ruled  paper,  find- 
ing it  to  cut  the  x-axis  at  x  =  14  and  the  y-axis  at  y  =  6. 

Look  now  for  the  corner  points  of  the  unit  squares  through  which 
this  straight  line  passes.  The  coordinates  of  these  points,  if  there  are 
such  points,  are  the  solutions  required.  In  this  case  the  line  passes 
through  only  one  such  point,  namely  the  point  (7,  3).  Hence  the 
solution  sought  is  x  =  7,  y  =  3. 

Ex.  2.     Find  the  least  positive  integers  which  satisfy 
7x-3y  =  17. 

Solution.  This  line  cuts  the  x-axis  at  x  =  2f  and  the  y-axis  at 
y  =  —  5|.  On  locating  these  points  as  accurately  as  possible,  the  line 
through  them  seems  to  cut  the  corner  points  (5,  6)  and  (8,  13).  The 
coordinates  of  both  these  points  satisfy  the  equation.  Hence  the 
solution  sought  is  x  =  5,  y  =  6. 

EXERCISES 

Solve  in  positive  integers  by  means  of  graphs,  and  check: 


1. 

x  +  y  =  7. 

5. 

90  -5 x  =  9y. 

2. 

x  +  y  =  3. 

6. 

5x=29-3y. 

3. 

x-27=-9y. 

7. 

140  -  7. r-  10,y  =  0. 

4. 

7  y  —  112  =  —  4  x. 

8. 

S-2x-  y  =  0. 

Solve  in  least  positive  integers,  and  check  : 
9.    7x  =  3y  +  21.  n.   4  x  =  9  y  —  36. 

10.    5x  -ky  =  20.  12.    5  x  -  2  y  +  10  =  0. 

68.  In  the  case  of  two  indeterminate  equations,  each  of  the 
first  degree  in  two  variables,  the  coordinates  of  the  point  of 
intersection  of  their  graphs  form  a  solution  of  both  equations. 


38  INTEGRAL   LINEAR  EQUATIONS 

Since  these  graphs  are  straight  lines,  they  have  only  one 
point  in  common,  and  hence  there  is  only  one  solution  of  the 
given  pair  of  equations. 

E.g.     On   graphing  x  +  y  =  4  and  y  —  x  —  2,  the  lines  are  found 

to  intersect  in  the  point  (1,  3).     Hence  the  solution  of  this  pair  of 

equations  is  .  Q 

x  —  1,  y  —  6. 

EXERCISES 

Graph  the  following  and  thus  find  the  solution  of  each  pair 
of  equations.     Check  by  substituting  in  the  equations. 

3x  —  2y  =  —2,  7     J8aj  =  7y, 

x  +  7  y  =  30.  1  x  +  3  =  5  y  +  3. 

aj  +  y  =  2,  [>/  =  !> 

3»  +  2?/  =  3.  '    l3//  +  4.r  =  y. 


[; 


x  —  4  y  =  1,  0     j  2  .r  —  4  ?/  =  4, 

a;  —  w  =  G  //  —  3. 


10.     '" 


J- 

l2s 


x  —  oy 


i.  "•  i,+.=; 


'    l.f-o  =  i/-l.  l3a;  +  2y=3. 

6.    \*  =  y-6'  12.    |-"'  =  72' 

1 5  y  =  x  +  9.  {  y  =  5. 

SOLUTION   BY   ELIMINATION 

69.  The  solution  of  a  pair  of  equations  such  as  the  foregoing 
may  be  obtained  without  the  use  of  the  graph  by  the  process 
called  elimination.     See  pages  154-159,  E.  C. 

70.  Elimination  by  substitution  consists  in  expressing  one 
variable  in  terms  of  the  other  in  one  equation  and  .substituting 
this  result  in  the  other  equation,  thus  obtaining  an  equation  in 
which  only  one  variable  appears.     See  §  116,  E.  C. 


SOLUTION  BY  ELIMINATION  39 

71.  Elimination  by  addition  or  subtraction  consists  in  making 
the  coefficients  of  one  variable  the  same  in  the  two  equations 
(§  62),  so  that  when  the  members  are  added  or  subtracted  this 
variable  will  not  appear  in  the  resulting  equation.  See  §  117, 
E.  C. 

72.  Elimination  by  comparison  is  a  third  method,  which  con- 
sists in  expressing  the  same  variable  in  terms  of  the  other  in  each 
equation  and  equating  these  two  expressions  to  each  other. 

As  an  example  of  elimination  by  comparison,  solve 

^y  +  x  =  U,  (1) 

(2) 
(3) 

(4) 


2y-  5a;  =-19 

From  (1), 

x  =  14  —  3  y. 

From  (2), 

5 

From  (3)  and 

W, 

14 

o         19  +  27/ 
3y  =              ». 
o 

y  =  3. 

Solving  (5), 

Substituting  in  (1), 

x  =  5. 

(5) 


Check  by  substituting  x  =  5,  y  =  3  in  both  (1)  and  (2). 

In  applying  any  method  of  elimination  it  is  desirable  first 
to  reduce  each  equation  to  the  standard  form :  ax  +-  by  =  c. 
See  §  119,  E.  C. 

EXERCISES 

Solve  the  following  pairs  of  equations  by  one  of  the  pro- 
cesses of  elimination. 

3  z +■  2  ?/ =  118,  \5x-8%  =  7y-U, 

x  +  5y  =  191.  \2x  =  y+%. 

3     (6x-3y  =  7,  \3x  +  7y-341  =  7},y  +  mx, 

\2x—2y  =  3.  '    l2U-+-J?/  =  l. 

5x-lly-2=±x,  [3?/  +  40  =  2.r+- 14, 


7. 


5  x  -  2 y  =  63.  19  y  -  347  =  5  x  -  420 

5y-3x  +  S  =  4y  +  2x  +  7,  f  6  y  —  5x  =  5  x  + 14, 

4x-2y=3y  +  2.  '    \3y-2x-6  =  5  +  x. 


40 


INTEGRAL   LINEAR   EQUATIONS 


f(»+5)(y+7)=(a?+l)(y-9)+112,     ±Q     \73-7y  =  ox, 


11. 


12. 


'  ax  =  6?/, 
.  x  +  2/  =  c. 


13 


la;—  ?/  =  b. 


14. 


f  a#  -\-by  =  c, 
Ia-  +  gy  =  h. 


15. : 


16/ 


.2^-305  =  12. 

f3  _  5  =  (. 
x     y 
2      3 

(SB       y 

-  +  -  =  c, 

x     y 

V+'i=h. 

I  ■'•    y 

SOLUTION   BY   FORMULA 

73.   We  now  proceed  to  a  more  general  study  of  a  pair  of 


linear  equations  in  two  variables. 

<2x  +  3y  =  4, 
\5x  +  6y  =  7. 


Ex.  1.    Solve 


Multiplying  (1)  by  5  and  (2)  by  2, 

5  •  2  x  +  5  •  3  y  —  5  •  4, 
2  •  5 x  +  2  •  6#  =  2  •  7. 

Subtracting  (3)  from  (4),   (2  •  6  -  5  •  ■'>)//  =  2.7  —  5-4. 


Solving  for  #, 


y 


2-6-5.3 


5  -  4_  -  I) 
3 


(1) 
(2) 

(3) 

(I) 
(•'») 

(6) 


In  like  manner,  solving  for  x  by  eliminating  y, 

4  •  6  -  7  •  3  _  J5_  _ 
2-6-5-3      -  3 


we  have 


-1. 


(7) 


Ex.   2.     In  this  manner,  solving, 

\7  x  +  9  y  =  71, 
12  a? +  3  y  =  48, 


we  find 


71  .  3  _  48  .  9        ,  7  -  48  -2-71 

-  and  y 


7  •  3  -  2  •  !> 


7-3-2-9 


*  Let  -  and      be  the  unknowns. 
x  y 


SOLUTION  BY  FORMULA  41 

In  Ex.  2,  the  various  coefficients  are  found  to  occupy  the 
same  relative  positions  in  the  expressions  for  x  and  y  as  the 
corresponding  coefficients  do  in  Ex.  1. 

Show  that  this  is  also  true  in  the  following: 

Ex.3,    pa- +  7// =  10,        Ex  4     $5x-3y  =  8, 
t2x  —  5  y  =  7.  12  a; +  7  y  =  19. 

A  convenient  rule  for  reading  directly  the  values  of  the  un- 
knowns in  such  a  pair  of  equations  may  be  made  from  the 
solution  of  the  following: 

Ex.  5.     Solve  -i    x     '     w        " 

{ a2x  +  b2y  =  c2. 

Eliminating  first  y  and  then  x  as  in  Ex.  1,  we  find: 

-  j, 

-2-1  and  y  = 


To  remember  these  results,  notice  that  the  coefficients  of  x  and  y  in 
the  given  equations  stand  in  the  form  of  a  square,  thus    1  A  and  that 

the  denominator  in  the  expressions  for  both  x  and  y  is  the  cross 
product  fljAo  minus  the  cross  product  a2bv  The  numerator  in  the 
expression  for  x  is  read  by  replacing  the  a's  in  this  square  by  the  c's, 

c   b 
i.e.,    l  ,  ,  and  then  reading  the  cross  products  as  before.     The  numera- 
c,  b2 

tor  for  y  is  read  by  replacing  the  b's  by  the  c's,  i.e.,    l    l,  and  then 

reading  the  cross  products.  2    2 


74.    To  indicate  that  the  coefficients  in  a  pair  of  equations  are 

=  axb2  —  a.2bx  and 


a  determinant.    These  are  much  used  in  higher  algebra. 


to  be  treated  as  just  described,  Ave  write     J    1 

\tto    Z>9 

call  ai  H 

a2  b2\ 

Since  any  pair  of  linear  equations  in  two  unknowns  may  be 
reduced  to  the  standard  form  as  given  in  Ex.  5,  it  follows  that 
the  values  of  x  and  y  there  obtained  constitute  a  form  ula  for 
tin-  solution  of  any  pair  of  such  equations. 


42  INTEGRAL   LINEAR   EQUATIONS 

EXERCISES 

Keduce  the  following  pairs  of  equations  to  the  standard  form 
and  write  out  the  solutions  by  the  formula: 
i      t3x  +  ±y  =  10,  6     jax-by  =  0, 

\4:X  +  y  =  9.  \x  —  y=c. 

1 4  x  —  5  y  =  —  26,  f  mas  +  ny  =  p, 

2*    \2x  —  3y=  -14.  '    lraj  +  sj/  =  i. 

f  (J  j/  -  17  =  -  5  a,  fa(a;  +  y) - b(x - y)=2a, 

l6a»-16=-5y.  '    \a(x-y)-b(x  +  y)=2b. 

fK*-S)=— Ky-2)+i^  o     J(fc  +  l>  +  (fc-2)y  =  3a, 
H(y-l)=aj-K*-2).         '    l(*+3)aj+(*  -4)</  = 

2  a;  -  y  =  53,  /  2  a.f  +  2%  =  4cr+  fc2, 

19  g  +  17  >/  =  0.  1  x  -  2  y  =  2  a  -  6. 

J  (a  +  b)x  —  (a  —  b)y  =  4  a&, 

1  (a  -  &)a?  +  (a  +  6)2/  =  2  a2  -  2  &2. 

m  (a  -  6)  -  i(a  -  3  &)  =  &  -  3, 
If  (a  -  6)  +  |(a  +  6)  =  18. 


3. 


(  a. 


11. 


12. 


13. 


a(«  +  y)  +  6(a?-y)  =  2, 


ta{x 

\a%x 


+  y)-  &"(»  - .'/)  =  «  -  &• 


,7(a;-5)  =3— 2  — as, 
14.    I    v  2 


15. 


( mx  +  ra.it/  =  ni:''  +  2  7/r'/;  +  ji8, 
t  nx  +  my  =  m3  +  2  inn2  +  n8. 

f  (w  +  a  \x  —  (m  —  w)t/  =  2  ?m, 
1  (m  +  Z)as  —  (in  —  V)y  =  2  win. 


17. 


19. 


|  :'.,  5  m  -  7  n  +  2)  -  fc(3  m  -  4  n  +  7)  =  w  +  3f, 
I  i'(  7  m  -  3  n  +  4)  -  ±(G  m  -  5  n  +  7)  =  w  -  2. 
r  (ft  +  fc)a,  +  (ft  _  jfe)y  =  2(//-  +  ft2), 
1  (ft  _  fc)as  +  (ft  +  Jc)y  =  2(ft2  -  F). 

i(a  +  &  _  c).r  +  Ka  -  6  +  n.v  =  a2  +  (b  -  c)2, 
i(a  —  6  +  c)x  +  \ (a  +  6  -  c)?/  =  a2  —  (6  —  c)\ 


INCONSISTENT  AND  DEPENDENT  EQUATIONS      43 


INCONSISTENT    AND    DEPENDENT    EQUATIONS 

75.  A  pair  of  linear  equations  in  two  variables  may  be  such 
that  they  either  have  no  solution  or  have  an  unlimited  number 
of  solutions. 

x-2y  =  -2,  (1) 

iSx-6y  =  -12,  (2) 


Ex.  1.     Solve 


On  graphing  these  equations  they  are  found  to  represent  two  par- 
allel lines.  Since  the  lines  have  no  point  in  common,  it  follows  that 
the  equations  have  no  solution.     See  Fig.  1. 

Attempting  to  solve  them  by  means  of  the  formula,  §  73,  we  find  : 

ar=(-2)(-6)-(-12)(-2)  =  -12 
l(-6)-3(-2)  0    ' 

l(-12)-3(-2)_  -6 
l(-6)-3(-2)        0  * 


and 


y  = 


12  —  6 

— -  and  are  not  numbers.      Hence,  from   this   it 

0  0 


But  by  §  25, 

follows  that  the  given  equations  have  no  solution.     In  this  case  no 
solution  is  possible,  and  the  equations  are  said  to  be  contradictory. 


>-i 

r 

•^ 

Ti 

\  v 

"X^ 

v> 

+i 

■---4 

<U>'        +1 

S 

V  ,],-;. 

^ 

i 

X 

+-        *"' 

Ai 

1 

_^0    , 

? 

5 

?!T 

:  V 

).0           :  1               :  - 

Ss 

-      y  l  '-lsr    -+- 

■  L 

ii^  - 

Fig.  1. 


Fig.  2. 


Ex.  2.     Solve 


r  3  x  —  6  y  =  —  6, 
\    x-2y  =  —  2, 


(1) 
(2) 


On  graphing  these  equations,  they  are  found  to  represent  the  same 
line.  Hence  every  pair  of  numbers  satisfying  one  equation  must 
satisfy  the  other  also.     See  Fig.  2. 


44  INTEGRAL   LINEAR   EQUATIONS 

Solving  these  equations  by  the  formula,  we  find  : 

x  =  (-6)(-2)-(-2)(-6)  =  0  and     =  3(-2)-l(-6)  =  0 
3(-2)-l(-6)  0  3(_2)-l(_6)      0 

But  by  §  24,  -   may  represent  any  number  whatever.     Hence  we  may 

select  for  one  of  the  unknowns  any  value  we  please  and  find  from  (1) 
or  (2)  a  corresponding  value  for  the  other,  but  we  may  not  select 
arbitrary  values  for  both  x  and  y. 

In  this  case  the  solution  is  indeterminate  and  the  equations 
are  dependent;   that  is,  one  may  be  derived  from  the  other. 
Thus,  (2)  is  derived  from  (1)  by  dividing  both  members  by  3. 

76.  Two  linear  equations  in  two  variables  which  have  one 
and  only  one  solution  are  called  independent  and  consistent. 

The  cases  in  which  such  pairs  of  equations  are  dependent  or  contra- 
dictory are  those  in  which  the  denominators  of  the  expressions  for  x 
and  y  become  zero.  Hence,  in  order  that  such  a  pair  of  equations 
may  have  a  unique  solution,  the  denominator  (1^.,  —  aJ^  of  the 
formula,  §  7-3,  must  not  reduce  to  zero.  This  maybe  used  as  a  test  to 
determine  whether  a  given  pair  of  equations  is  independent  and  con- 
sistent. 

EXERCISES 

In  the  following,  show  both  by  the  formula  and  by  the  graph 
which  pairs  of  equations  are  independent  and  consistent,  which 
dependent,  and  which  contradictory. 

|  5  x  —  3  y  =  5,  \  3x— 6  y+5  =  2  x — 5  y +7, 

I  5  a  —  3?/ =  9.  '    iox+3!/  —  l  =  3x  +  5y+3. 

tx-7  +  5y  =  y-x-2,  2y  +  7x  =  2  +  6x, 

5x  +  3y  —  4:  =  3x  —  y+3.      '    U»-3y  =  4  +  3x-5y. 

3  j7x-3y-4  =  2x-2,  g    (5as-3  =  7y  +  8, 

\x  +  y  —  3=2x  —  7.  '    L2a5  +  7=4y  — 9. 

4  (X-3y  =  G,  (5x  +  2y  =  6  +  3x  +  5y, 
1 5  x  — 15^  =  18.                         '    l3x  +  y  =  18  —  3x  +  10y. 

g       3y-4x-l  =  2x-5y+$,1Q     \3x  +  4y  =  7  +  5y, 
i  2  y  —  5  x  +  8  =  3  x  +  //.  x  —y  =  6  —  2  x. 


EQUATIONS  IN  MORE   THAN    TWO    VARIABLES      45 
SYSTEMS  OF   EQUATIONS   IN  MORE    THAN   TWO   VARIABLES 

77.  If  a  single  linear  equation  in  three  or  more  variables  is 
given,  there  is  no  limit  to  the  number  of  sets  of  values  which 
satisfy  it. 

E.g.  3  x  +  2  y  +  1  z  =  21  is  satisfied  by  x  =  1,  y  =  3,  z  =  3f ;  x  =  2, 
y  =  2,  z  =  3£ ;  x  =  0,  y  =  0,  2  =  6;  etc. 

If  two  linear  equations  in  three  or  more  variables  are  given, 
they  have  in  general  an  unlimited  number  of  solutions. 

E.g.  3  x  +  2  y  +  I  z  =  21  and  a;  +  y  +  z  =  6  are  both  satisfied  by 
x  =  2,  y  =  —  1,  2  =  5 ;  a;  =  3,  y  =  —  l£,  2  =  4^ ;  etc. 

But  if  a  system  of  linear  equations  contains  as  many  equa- 
tions as  variables,  it  has  in  general  one  and  only  one  set  of 
values  which  satisfy  all  the  equations. 

fx  +  y  +  2  =  6, 
E.g.  The  system   -j  3  x  —  y  -\-  2  z  =  7, 
[2  x  +  3  y  -  z  =  5, 
is  satisfied  by  a;  =  1,  y  =  2,  2  =  3,  and  by  no  other  set  of  values. 

It  may  happen,  however,  as  in  the  case  of  two  variables, 
that  such  a  system  is  not  independent  and  consistent. 

Such  cases  frequently  occur  in  higher  work,  and  a  general  rule  is 
there  found  for  determining  the  nature  of  a  system  of  linear  equa- 
tions without  solving  them  :  namely,  by  means  of  determinants  (§  7:5).  In 
this  book  the  only  test  used  is  the  result  of  the  solution  itself  as 
explained  in  the  next  paragraph. 

78.  An  independent  and  consistent  system  of  linear  equa- 
tions in  three  variables  may  be  solved  as  follows : 

From  two  of  the  equations,  say  the  1st  and  2d,  eliminate  one  of 
the  variables,  obtaining  one  equation  in  the  remaining  two  variables. 

From  the  1st  and  3d  equations  eliminate  the  same  variable,  ob- 
taining a  second  equation  in  the  remaining  two  variables. 

Solve  as  usual  the  two  equations  thus  found.  Substitute  the 
values  of  these  two  variables  in  one  of  the  given  equations,  and  thus 
find  the  value  of  the  third  variable. 

The  process  of  elimination  by  addition  or  subtraction  is  usually 
most  convenient.     See  §  120,  E.  C. 


46 


INTEGRAL   LINEAR   EQUATIONS 


EXERCISES 


Solve  each  of  the  following  systems  and  check  by  substitut- 
ing in  each  equation  : 


x  +  5  y  —  i 


2  =  9, 


5  x  —  y  +  3  z  =  16, 

[T  x  -\-  6  y  A-  z  =  34. 


r8  z  -  3  y  +  x  =  -  2, 
3  x  —  5  y  —  6  z  =  —  46, 

[y  +  o  x  —  4  z  =  —  18. 


[a  +  6  +  c  =  9, 
8a  +  46  +  2c  =  36, 
.27  a  +  96+3c  =  98. 


*  Use    ,  7,  and  -  as  the  unknowns. 
a  b  c 


f3  =  2 
a      b 


6.*  J 


-  +  r--  =  17, 
a      b      c 


3     6 
6     c 


(18  I  —  7  wi  —  5  7i ,  =  161, 
1 1  in.  —  |  Z  +  n  =  18, 
31  h  +  2  m  +  f  Z  =  33. 


7. 


(x  +  2y-3z  = 

>  2  x  —  3  ?/  +  z  = 
>  .r  —  4  y  —  7  z 


:i  +  (i+9 


4      i«  +  &  +  _c_=_4 
<  6  ^  9  ^  12 


9     12     15 


fa;  +  »/  =  16, 

9.      z  +  .r  =  22, 

U  +  2  =  28. 

f.r  +  2y  =  26, 
10.    j3aj  +  4z  =  56, 
[5  y  +  6  2  =  65. 


[2.T  +  3>/-7z  =  19, 

8.      5ar  +  82/  +  llz=24, 
U  x  +  lly  +  4z=  43. 

Show  that  this  system  is  not 
independent. 


ra  +  6  +  c  =  5, 
11.    Is  a  —  5  b  +  7  c  =  79, 
9  a  -  11  b  =  91. 


f  I  +  «i  +  n  =  29£, 

12.      /  +  ///-  ra  =  18i 
[l-  m+  n  =  L3f. 


EQUATIONS   IN  MOBE   THAN   TWO    VARIABLES      47 


13. 


14. 


I  +  m  +  n  =  a, 

,  /  +  m  —  n  =  b, 
[l  —  m  +  n  =  c. 

r7/  +  clz  =  q, 
[ex  +fz  =  r. 


15. 


a      o 

=  4, 

a      c 

=  3, 

7  +  l- 

b      c 

=  2. 

16.     I 


17. 


u  +  2v  +  3x  +  4y  =  30, 

2  u  +  3  v  +  4  x  +  5  y  =  40, 

3  «  +  4  i>  +  5  .«  +  6  y  =  50, 

4  u  +  5  v  +  6  x  +  7  ^/  =  60. 


=  a, 

=  &, 

2/      2 

=  c. 

18.    Make  a  rule  for  solving  a  system  of  four  or  more  linear 
equations  in  as  many  variables  as  equations. 


PROBLEMS   INVOLVING  TWO   OR  MORE   UNKNOWNS 

1.  A  man  invests  a  certain  amount  of  money  at  4%  inter- 
est and  another  amount  at  5%,  thereby  obtaining  an  annual 
income  of  $3100.  If  the  first  amount  had  been  invested  at 
5%  and  the  second  at  4%,  the  income  would  have  been 
$3200.     Find  each  investment. 

2.  The  relation  between  the  readings  of  the  Centigrade 
and  the  Fahrenheit  thermometers  is  given  by  the  equation 
F  =  32  +  f-  C.  The  Fahrenheit  reading  at  the  melting  tem- 
perature of  osmium  is  2432  degrees  higher  than  the  Centigrade. 
Find  the  melting  temperature  in  each  scale. 

In  the  Reaumur  thermometer  the  freezing  and  boiling  points  are 
marked  0°  and  80°  respectively.  Hence  if  C  is  the  Centigrade  reading 
and  R  the  Reaumur  reading,  then  R  =  ^  C.     See  §  101,  E.  C. 

3.  "What  is  the  temperature  Fahrenheit  (a)  if  the  Fahren- 
heit reading  equals  ^  of  the  sum  of  the  other  two,  (b)  if  the 
Centigrade  reading  equals  |  of  the  Fahrenheit  minus  the  Reau- 
mur, (c)  if  the  Reaumur  is  equal  to  the  sum  of  the  Fahrenheit 
and  Centigrade  ? 


48  INTEGRAL   LINEAR    EQUATIONS 

4.  Going  with  a  current  a  steamer  makes  19  miles  per 
hour,  while  going  against  a  current  -|  as  strong  the  boat  makes 
5  miles  per  hour.     Find  the  speed  of  each  current  and  the  boat. 

5.  There  is  a  number  consisting  of  3  digits  whose  sum  is  1 1. 
If  the  digits  are  written  in  reverse  order,  the  resulting  number 
is  594  less  than  the  original  number.  Three  times  the  tens'  digit 
is  one  more  than  the  sum  of  the  hundreds'  and  the  units'  digit. 

6.  A  certain  kind  of  wine  contains  20  %  alcohol  and  another 
kind  contains  28%.  How  many  gallons  of  each  must  be  used 
to  form  50  gallons  of  a  mixture  containing  21.6  r/0  alcohol  ? 

7.  The  area  of  a  certain  trapezoid  of  altitude  8  is  68.  If  4 
is  added  to  the  lower  base  and  the  upper  base  is  doubled,  the 
area  is  108.     Find  both  bases. 

A  trapezoid  is  a  four-sided  figure  whose  upper  base  bv  and  lower 
base,  b„,  are  parallel,  but  the  other  two  sides  are  not.  If  h  is  the  per- 
pendicular distance  between  the  bases,  then  the  area  is  a  =  -  (/>r  +  b„). 

8.  If  on  her  second  westward  journey  the  Lusitania  had 
made  1  knot  more  per  hour,  she  w7ould  have  crossed  in  4 
hours  and  38  minutes  less  than  she  did.  But  if  her  speed  had 
been  4  knots  per  hour  less,  she  would  have  required  23  hours 
ami  10  minutes  longer.  Find  the  time  of  her  passage  and  her 
average  speed  if  the  length  of  her  course  was  2780  knots. 

9.  Aluminium  bronze  is  an  alloy  of  aluminium  and  copper. 
The  densities  of  aluminium,  copper,  and  aluminium  bronze  are 
2.6,  8.9,  and  7.69  respectively.  How  many  com.  of  each 
metal  arc  used  in  100  ccm.  of  the  alloy  ?     See  §  99,  E.  C. 

10.  Wood's  metal,  which  is  used  in  fire  extinguishers  on 
account  of  its  low  melting  temperature,  is  an  alloy  of  bismuth, 
lead,  tin,  and  cadmium.  In  120  pounds  of  Wood's  metal,  -|  of 
the  tin  plus  '.  of  the  lead  minus  .,1„  of  the  bismuth  equals 
7  pounds.  If  \  of  the  lead  and  |  of  the  tin  be  subtracted 
from  the  bismuth,  the  remainder  is  ll'  pounds.  Find  the 
amount  of  each  metal  if  15  pounds  of  cadmium  is  used. 


PROBLEMS  IN   TWO    OR   MORE    VARIABLES  49 

11.  The  upper  base  of  a  trapezoid  is  G  and  its  area  is  168. 
If  i  the  lower  base  is  added  to  the  upper,  the  area  is  210. 
Find  the  altitude  and  the  lower  base. 

12.  A  and  B  can  do  a  piece  of  work  in  18  days,  B  and  C  in 
24  days,  and  C  and  A  in  36  days.  How  long  will  it  require 
each  man,  working  alone,  to  do  it,  and  how  long  will  it  require 
all  working  together  ? 

13.  A  and  B  can  do  a  piece  of  work  in  m  days,  B  and  C  in  n 
days,  and  C  and  A  in  p  days.  How  long  will  it  require  each  to 
do  it  working  alone? 

14.  A  beam  resting  on  a  fulcrum  balances  when  it  carries 
weights  of  100  and  130  pounds  at  its  respective  ends.  The 
beam  will  also  balance  if  it  carries  weights  of  80  and  110 
pounds  respectively  2  feet  from  the  ends.  Find  the  distance 
from  the  fulcrum  to  the  ends  of  the  beam. 

15.  A  beam  carries  three  weights,  A,  B,  and  C.  A  balance 
is  obtained  when  A  is  12  feet  from  the  fulcrum,  B  8  feet  from 
the  fulcrum  (on  the  same  side  as  A),  and  C  20  feet  from  the 
fulcrum  (on  the  side  opposite  A).  It  also  balances  when  the 
distance  of  A  is  8  feet,  B  10  feet,  and  O  18  feet.  Find 
the  weights  B  and  C  if  A  is  50  lbs. 

16.  At  0°  Centigrade  sound  travels  1115  feet  per  second 
with  the  wind  on  a  certain  day,  and  1065  feet  per  second 
against  the  wind.  Find  the  velocity  of  sound  in  calm  weather, 
and  the  velocity  of  the  wind  on  this  occasion. 

17.  If  the  velocities  of  sound  in  air,  brass,  and  iron  at  0° 
Centigrade  are  x,  y,  z  meters  per  second  respectively,  then 
3x+2y—  z  =  2505,  5x  —  2y  +  z  =  151,  and  x  +  y  +  z  =  8777. 
Find  the  velocity  in  each. 

18.  If  x,  y,  z  are  the  Centigrade  readings  at  the  temperatures 
which  liquefy  hydrogen,  nitrogen,  and  oxygen  respectively, 
then  3 x-Sy+2z=  440,  -  8 x  +  2  ?/  +  4  z  =  903,  and  x  +  4 y 
—  62  =  60.  Find  each  temperature  in  both  Centigrade  and 
Fahrenheit  readings. 


50  INTEGRAL   LINEAR   EQUATIONS 

19.  Two  trapezoids  have  a  common  lower  base.  Their  alti- 
tudes are  8  and  10  respectively,  and  the  sum  of  their  areas  is 
148.  If  the  upper  base  of  the  first  trapezoid  is  multiplied  by  2 
and  that  of  the  second  divided  by  2,  their  combined  area  is 
ir»i' ;  while  if  the  upper  base  of  the  first  is  divided  by  2  and 
that  of  the  second  multiplied  by  2,  the  combined  area  is  17<<. 
Find  the  bases  of  each  trapezoid. 

20.  If  x,  y,  z  are  the  Centigrade  readings  at  the  freezing 
temperatures  of  hydrogen,  nitrogen,  and  oxygen  respectively, 
then  we  have  x  +  y  —  3  z  =  199,  2  x  —  5  y  +  z  =  328,  and 
—  4  x  +  2  y  +  2  z  =  156.     Find  each  temperature. 

21.  If  x,  y,  z  are  respectively  the  melting  point  of  carbon. 
the  temperature  of  the  hydrogen  flame  in  air,  and  the  tempera- 
ture of  this  flame  in  pure  oxygen,  then  10  x  -+-  2  y  +  z  =  41,892, 
15 x _|_ y  +  2 z  =60,212,  and  7  x  ±y  +  z  =  29,368.     Find  each. 

22.  If  a,  b,  c  are  the  values  in  millions  of  the  mineral 
products  of  the  United  States  in  1880,  1900,  and  1906  re- 
spectively, find  each  from  the  following  relations  : 

k         b       c       i  — ()    a  .  b   .  c      rrn  b  .  c 

5  a =  1oi2,  -  -\ h  -  =  669,  a \-  -  =  37. 

8      10  '345  2     7 

23.  If  x,  y,  z  represent  in  thousands  of  tons  the  steel 
products  of  the  United  States  in  1880,  1890,  and  1905,  find 
each  from  the  following  relations : 

x  +  y  +  z  =  25,547,  3x  +  iy-z  =  826,  x -3y  +  z  =  8  139. 

24.  If  the  number  of  millions  of  tons  of  coal  mined  in  the 
United  States  in  1890,  L900,  and  L906  be  represented  by  x,  y,  z 
respectively,  find  each  from  the  following  relations: 

E  +  1L  +  A  =1 05,  x-  v  +  —  =  153,  3  x  +  2  y  -  2 z  =  164. 
2     30      25  9     17 

25.  If  the  values  in  millions  of  the  farm  products  of  the 
United  States  in  1870,  L90<>,  and  190<;  arc  represented  by  1.  m. 
and  n  respectively,  find  each  from  the  following  relations : 

2 1  +  m  - n  =  1633,  3 1  —  2 m  +  n=  3440,  I  +  in  +  n  =  13,675. 


PROBLEMS  IN   TWO   OR  MORE   VARIABLES  51 

26.  The  sum  of  the  areas  of  two  trapezoids  whose  altitudes 
are  10  and  12  respectively  is  284.  If  the  upper  base  of  the 
first  is  multiplied  by  3  while  its  lower  is  decreased  by  2,  and 
the  upper  base  of  the  second  is  divided  by  2  while  its  lower 
base  is  increased  by  3,  the  sum  of  the  areas  is  382 ;  if  the 
upper  bases  of  both  are  doubled  and  the  lower  bases  of  both 
divided  by  2,  the  sum  of  the  areas  is  322;  and  if  the  upper- 
bases  are  divided  by  2  while  the  first  lower  is  doubled  and  the 
second  trebled,  the  sum  of  the  areas  is  388.  Find  the  bases  of 
each  trapezoid. 

27.  Two  boys  carry  a  120-pound  weight  by  means  of  a  pole, 
at  a  certain  point  of  which  the  weight  is  hung.  One  boy 
holds  the  pole  5  ft.  from  the  weight  and  the  other  3  ft. 
from  it.     What  proportion  of  the  weight  does  each  boy  lift  ? 

Solution.  Let  x  and  y  be  the  required  amounts,  then  5x  is  the 
leverage  of  the  first  boy  and  3  y  that  of  the  second,  and  these  must  be 
equal  as  in  the  case  of  the  teeter,  p.  122,  E.  C.     Hence  we  have 

5  x  =  3  y,  and  x  +  y  =  120. 

Solving,  we  find  x  =  45,  y  =  75. 

28.  If,  in  problem  27,  the  boys  lift  Px  and  P2  pounds  respec- 
tively at  distances  dx  and  d2,  and  w  is  the  weight  lifted,  then 

PA  =  P.A,  (1) 

P1  +  P2  =  w.  (2) 

Solve  (1)  and  (2),  (a)  when  Px  andP2  are  unknown,  (b)  when 
Pj  and  w  are  unknown,  (c)  when  Px  and  d2  are  unknown. 

29.  A  weight  of  540  pounds  is  carried  on  a  pole  by  two  men  at 
distances  of  4  and  5  feet  respectively.    How  much  does  each  lift  ? 

30.  A  weight  of  470  pounds  is  carried  by  two  men,  one  at  a 
distance  of  3  feet  and  the  other  lifting  200  pounds.  At  what 
distance  is  the  latter  ? 

31.  Two  men  are  carrying  a  weight  on  a  pole  at  distances  of 
4  and  6  feet  respectively.  The  former  lifts  240  pounds.  How 
many  pounds  are  they  carrying? 


CHAPTER   V 
FACTORING 

79.  A  rational  integral  expression  is  said  to  be  completely 
factored  when  it  cannot  be  further  resolved  into  factors  which 
are  rational  and  integral.     Such  factors  are  called  prime  factors. 

The  simpler  forms  of  factoring  are  given  in  the  following 
outline. 

A  monomial  factor  of  any  expression  is  evident  at  sight, 
and  its  removal  should  be  the  first  step  in  every  case. 

E.g.  4  ax1  +  2  a-x  =  2  ax(2  x  +  a). 

FACTORS   OF   BINOMIALS 

80.  The  difference  of  two  squares. 

E.g.  4  x2  -  9  z*  =  (2  x  +  3  z2)  (2  x  -  3  z2) . 

81.  The  difference  of  two  cubes. 

E.g.     8  a?  -  27  f  =  (2  x  -  3  y)  [  (2  x)  2  +  (  2  x)  (3  //)  +  (3  y) 2] 
=  (2  x  -  3  //)  ( 1  .r-  +  6  xy  +  9  //-). 

82.  The  sum  of  tv:o  cubes. 

E.g.      27  z8  +  6 1  f  =  (3  x  +  4  y) [(3  sc)2-  (:*>  x)  (  1  //)  +  (1  //)'-] 
=  (3a;  +  l//)(!i./-:  -  12 xy  +  10  if). 

FACTORS  OF  TRINOMIALS 

83.  Trinomial  squares. 

E.g.  a2  +  2  ab  +  ft2  =  (o  +  6)2  =  (a  +  &)(a  +  A), 

ami  a2  —  2  afi  +  62  =  (a  —  /;)2  =  (a  —  6)(a  —  /;). 

52 


FACTORS   OF  TRINOMIALS  53 

84.  Trinomials  of  the  form  x2  -\-px  +  q. 

E.g.  x2  +  3  x  -  10  =  (x  +  5)  (x  -  2). 

A  trinomial  of  this  form  has  two  binomial  factors,  x  +  a  and  x  +  b, 
if  two  numbers  a  and  b  can  be  found  whose  product  is  q  and  whose 
algebraic  sum  is  p. 

85.  Trinomials  of  the  form  mx2  +  nx  +  r. 

£.<7.  6  a:2  +  7  j;  -  20  =  (3  x  -  4)(2  x  +  5). 

A  trinomial  of  this  form  has  two  binomial  factors  of  the  type 
ax  +  6  and  ex  +  ^/,  if  four  numbers,  a,  b,  c,  d,  can  be  found  such  that 
ac  =  m,  bd  —  r,  and  ad  +  be  =  n.     See  §  142,  E.  C. 

86.  Trinomials  which  reduce  to  the  difference  of  two  squares. 

E.g.     x4  +  x2y2  +  if  =  x4  +  2  x2y2  +  v/4  -  x2y2  =  (x2  +  y-)2  -  x2y'2 
=  (x2  +  if  -  xy)(x2  +  y-  +  xy). 

In  this  case  xhf  is  both  added  to  and  subtracted  from  the  expres- 
sion, whereby  it  becomes  the  difference  of  two  squares.  Evidently 
the  term  added  and  subtracted  must  itself  be  a  square,  and  hence  the 
degree  of  the  trinomial  must  be  4  or  a  multiple  of  4,  since  the  degree 
of  the  middle  term  is  half  that  of  the  trinomial. 

Ex.         4  a8  -  16  a4Z/4  +  9  bs  =  4  as  -  12  aHA  +  9  bs  -  4  a4b* 
=  (2  a4  -  3  ft4)'2  -  4  a*h* 
=  (2  a4  -  3  b*  +  2  a2b2)  (2  a4  -  3  ft4  -  2  aa6a). 

EXERCISES  ON  BINOMIALS  AND  TRINOMIALS 

Factor  the  following : 

1.  as  +  b3.                    5.  7  ace2  -  56  aV.  9.  |  ^  — -^  rs2. 

2.  a3  —  63.                    6.  a5-a64.  10.  8r4—  27  r. 

3.  (a  +  6)8-c8.          7.  121  a7 -4 as/4.  11-  (a  +  &)2-c2. 

4.  (a  +  &)8  +  c*.           8.  £a8  +  Tk6'-  12-  c2-(a-&)2. 

13.  5  c2  +  7  erf  —  6  d2.  15.    4 a;2  —  12  a$  +  9  ?/2. 

14.  a?4  —  3afy*  +  y*.  16.   a?2  + 11  ax?  +  30  z2. 


54  FACTORING 

17.  G  .?•'-  —  5 xy  —  G  y2.  24.  a2  4- 10  a  —  39. 

18.  3  aVy4 - 69  a2a«/24-  336  a2.  25.  8  afy3  -  48  a2y2z  +  72  a2*/*2. 

19.  20a262  +  23a&a;-2l!B2.  26.  4  m8  -  60  mV  +  81  n*. 

20.  a4  +  2  a2b2  +  9  64.  27.  35  a2*  -  6  aV-  9  ft2*. 

21.  48  crViy  -  75  a/.  28.  (a  +  bf—  (c  —  d  f. 

22.  16  a4.r//  +  5  I  ay4.  29.  72  aW  -  19  aay2  -  40  ,y4. 

23.  aj42/2  +  2a;22/«4-22.  30.  4(a-3)6-3762(a-3)3+964. 

31.  6(s  +  y)«+6(a*-y»)-6(a!-y)» 

32.  9(as  -a)2-  24(aj  -  a)  (as  +  a)  +  16(as  +  «)2. 

33.  12(c  +  rf)2-  7(c  +  d)(c-d)— 12(c  — d)1. 

34.  (a2  +  5  a  —  3)2-  25(a2  +  5  a  —  3)  +  150. 

FACTORS  OF  POLYNOMIALS  OF  FOUR  TERMS 

A  polynomial  of  four  terms  may  be  readily  factored  in  case 
it  is  in  any  one  of  the  forms  given  in  the  next  three  paragraphs. 

87.  It  may  be  the  cube  of  a  binomial. 
Ex.1,     a3  -  3  a?b  +  3ab"--b3  =  (a  -  bf. 
Ex.  2.     8  x*  4-  36  x2y  +  54  xtf  +  27  f 

=  (2x)«  +  3(2x)*(3*/)  +  8(2x)(8y)»  +  (3y)« 
=  (2x  +  3y)8.     See  Ex.  34,  (rf),  p.  23. 

88.  It  may  be  resolvable  into  the  difference  of  two  squares. 
In  this  case  three  of  the  terms  must  form  a  trinomial  square. 
Ex.  1.     a2  -  c2  +  2  a&  +  &2  =  (a2  +  2  a&  +  &2)  -  c2 

=  (a  +  &)2  -  c2  =  (a  +  6  +  c)(a  +  &  -  c). 

Ex.  2.     4  a;2  +  zc'  -  4  a'4  -  1  =  zG  -  4  x*  +  4  .jr  -  1 

=  z6  -  (4x-4  -  4x2  +  1)  =  z&  -  (2x2  -  I)2 

=  (;,  +  2x2-l)(c:i-2/-+  1). 


FACTORS  FOUND   BY  GROUPING  55 

89.  A  binomial  factor  may  be  shown  by  grouping  the  terms. 

In  this  case  the  terms  are  grouped  by  twos  as  in  the  following 
examples. 

Ex.  1.   ax  4-  ay  +  bx2  4-  bxy  =  (ax  4-  ay)  4-  (bx2  4-  bxy) 
=  a(x  +  y)  +  ix(x  +  y)  =  (a  +  for)  (a:  +  y). 

Ex.  2.   aa;  +  6oj  +  a2  —  62  =  (ax  +  6a)  +  (a2  -  b2) 

-  x(a  +  ft)  +  (a  -  ft)(a  +  ft)  =  (x  +  a  -  ft)(a  +  ft). 

EXERCISES 

Factor  the  following  polynomials: 

1.  a'3  +  3  x':y  4-  3  cc?/2  +  2/3.  8.  a2b2  —  a2bc2n  —  abn  4-  tm2c2. 

2.  8 a3 - 36 a2b  +  54 ab2 - 27 63.      9.  2y2  +  <Lby  +  3cy  +  6bc. 

3.  4  a4  —  4  a262  +  64  —  16  x2.  10.  bcyz  +  c2z2  +  bdy  +  dcz. 

4.  2  aa"  4- 3  6c  4- 2  ac  4- 3  6a\  11.  5a2c  +  12ca"  — 6  ad— 10  ac2. 

5.  27  x3  —  54  a%  4-  36  xy2  —  8  ?/3.  12.  a2  —  62.i'2  +  acx2  —  bcx5. 

6.  36  a4 -24  a3 +  24  a -16.  13.  bz c2  -  chf  -  Ihf  +  y5 . 

7-  mx.v2  —  mrx  —  rn2x  +  /•'-'// .  14.  a3*  —  2  a2*6*  —  2  a*624  4-  b3k. 

15.  ma+i  +  ma»"  4-  mbnb  +  //"f6. 

1 6.  b2[f  —  b(c  —  d)  y2  4-  d  (by  -  c)  4-  d2. 

FACTORS  FOUND  BY  GROUPING 

90.  The  discovery  of  factors  by  the  proper  grouping  of  terms 
as  in  §  89  is  of  wide  application.  Polynomials  of  five,  six,  or 
more  terms  may  frequently  be  thus  resolved  into  factors. 

Ex.  1.    a2  4-  2  ab  4-  b2  4-  5  a  4-  5  b  =  (a  4-  6)2  4-  (5  a  4-56) 

=  (a  +  ft)  (a  +  ft)  +  5(a  +  ft)  =  («  +  b  +  5)  (a  +  ft). 
Ex.  2.   x2  —  7  a;  +  6  —  ax  4-  6  a  =  or  —  7  x  +  6  —  (aa;  —  6a) 

=  (.r-  l)(x  -  6)-  a(x-  6)  =  (x  -  6)  (a;  -  1  -a). 
Ex.  3.    a2  -  2  a6  +  62  -  a2  -  2 .17/  -  ?/2 

=  (a2  -  2  aft  +  ft2)  -  (r-  +  2  .»•//  +  //-) 

=  (a  -  ft)2  -  (a  -  y)2  =  (a  -  b  +  x  -  y)  (a  -  ft  -  x  +  y) . 


56  FACTORING 

Ex.  4.    ax2  4-  ax  —  6  a  +  sc2  4  ~  ■>"  +  12 

=  «(./•-  +  x  -  6)  +  (./-  4  7x4  12) 
=  a(z  +  3)0  -  2)  +  (x  +  3)(.r  +  4) 

=  (•'•  +  •"')  L"  ( ■'•  --')+  ->-  +  4]  =  (-r  +  3)  (a*  -  2  a  +  x  +  4) . 
In  some  cases  the  grouping  is  effective  only  after  a  term  has 
been  separated  into  two  parts. 

Ex.  5.   2  a3  +  3  a2  +  3  a .  +  1  =  a?  +  (a8  +  3  a2  +  3  a  + 1) 

=  a34(a  +  l)8  =  (a  +  a+l)[a2-a(a  +  l)  +  (a  +  1)-] 
=  (2a  +  l)(a2  +  a  +  l). 

As  soon  as  the  term  2  o3  is  separated  into  two  terms  the  expression  is 
shown  to  be  the  sum  of  two  cubes. 

Again,  the  grouping  may  be  effective  after  a  term  has  been 
both  added  and  subtracted : 

Ex.  G.   a4  +  1/  =  (a4  +  2  a262  +b*)-2  a2b2 
=  (a2  +  62)2-(a&V2)2 
=  (a2  +  b-  +  ab  \/2)(a2  +  //-  -  a&V2). 

Tn  tliis  ease  the  factors  arc  irrational  as  to  one  coefficient.     Such 
factors  are  often  useful  in  higher  mathematical  work. 

EXERCISES 

Factor  the  following : 

1.  x2  —  2  xy  +  y2  —  ax  +  ay.  3.  a8  —  6s  —  a2  —  a6  —  ft2. 

2.  a2  —  a&  +  62  +  a3  4-  68.  4.   a-  —  2ab+b2—  x2+2xy— y2. 

5.  a4  +  2  as6  —  a2c2  +  a2&2  —  2  a&c2  —  ft2©2. 

6.  .r4  -  //4  +  aa;2  +  ay2  —  -r  —  //'-'.    7.   a4  +  a2b2  +  b4  +  a3  4-  &3. 
In  7  group  the  first  three  and  the  last  two  terms. 

8.  a3  —  1  4-  3  x  —  3ar  +  3r3-     ( I  roup  the  last  four  terms. 

9.  xs  4-  as2  4-  3  x  +  y3  —  y2  4-  3  y. 

Gron;>  in  pairs,  the  1st  and  1th.  2d  and  5th,  3d  and  6th  terms. 

10.  x*  4-  x*y  —  xy3  —  y4  4-  •',:'  —  y3. 

1 1 .  a4  +  -1  <fb  +  6  ((-//-  4-  4  ab"  +  b4-  x*. 


THE  FACTOR    THEOREM  57 

12.  x4  +  4  x2z  —  4  y-  +  4  ?/zo  +  4  z2  —  to2. 

13.  2  a2  - 12  b2  +  3  6c?  -  5  aft  -  9  6c-  6  ac  +  2ad. 
Group  the  terms :   2  a2  —  5  ai  —  12  62. 

14.  a2  +  a6  -  4  ac  —  2  62  +  4  be  +  3  ad  -  3  bd. 

15.  a3  4-  2  -  3  a,     Group  thus :  (a3  -  1)  +  (3  -  3  a). 

16.  4  a2  +  a  —  8  ax  —  a;  +  4  a;2. 

17.  3  a2  —  8  ab  +  4  b2  +  2  ac  —  4  6c. 

18.  £8  +  ?/s,  also  a;16  +  ?/1<;. 

19.  aG  +  2  a363  +  b6  -  2  a46  -  2  a,64. 

20.  a3  -  3  a2  +  4.      Group  thus  :  (a3  -  2  a2)  +  (4  -  a2). 

21.  a2c  —  ac2  —  a26  +  ab2  —  62c  4-  6c2. 

22.  orb  —  a2c  +  b2c  —  a62  4-  ac2  —  6c2. 

23.  3  Xs  —  x2  —  4  x-  4-  2.      Add  and  subtract   -  2  x2. 

24.  2  x3  —  11  x2  + 18  x  -  9.      Add  and  subtract  9  x2. 

FACTORS    FOUND    BY    THE    FACTOR    THEOREM 

91.   It  is  possible  to  determine  in  advance  whether  a  poly- 
nomial in  x  is  divisible  by  a  binomial  of  the  form  x  —  a. 

E.g.   In  dividing  x4  —  4  x3  +  7  x-  —  7  x  +  2  by  x  —  2,  the  quotient  is 
found  to  be  x3  —  2  x2  +  3  x  —  1. 

Since  Quotient  x  Divisor  =  Dividend,  we  have 

(x-2)(x3-2x2  +  3x-  l)  =  x4  -4x3  +  7x2-7x  +  2. 

As  this  is  an  identity,  it  holds  for  all  values  of  x.     For  x  =  2  the  fac- 
tor (x  —  2)  is  zero,  and  hence  the  left  member  is  zero,  §  22. 

Hence  for  x  =  2  the  right  member  must  also  be  zero.     This  is  in- 
deed the  case,  viz. : 

04  _  4  .  03  +  7  .  22  -  7  •  2  +  2  =  16  -  32  +  28  -  14  +  2  =  0. 

Hence  if  x  -  2    is  a  factor  of  x4  -  4  xs  +  7  x2  -  7  x  +  2,  the  latter 
must  reduce  to  zero  for  x  =  2. 


58  FACTORING 

92.  Theorem.  If  a  polynomial  in  x  reduces  to  zero  when 
a  particular  number  a  is  substituted  for  x,  thru  x  —  a  is  a 
factor  of  the  polynomial,  and  if  the  substitution  of  a  for 
x  does  not  reduce  the  polynomial  to  zero,  then  x  —  a  is  not 
a  factor. 

Proof.  Let  D  represent  any  polynomial  in  x.  Suppose  D  lias  been 
divided  by  x  —  a  until  the  remainder  no  longer  contains  x.  Then, 
calling  the  quotient  Q  and  the  remainder  R,  we  have  the  identity 

D=(2(x-a)+R,  (1) 

which  holds  for  all  values  of  x. 

The  substitution  of  a  for  x  in  (1)  does  not  affect  R,  reduces 
Q(x  —  o)  to  zero,  and  may  or  may  not  reduce  D  to  zero. 

(1)  If  x  =  a  reduces  D  to  zero,  then  0  =  0  +  R.  Hence  R  is  zero, 
and  the  division  is  exact.     That  is,  x  —  a  is  a  factor  of  D. 

(2)  If  x  =  a  does  not  reduce  D  to  zero,  then  R  is  not  zero,  and  the 
division  is  not  exact.     That  is,  x  —  a  is  not  a  factor  of  D. 

93.  In  applying  the  factor  theorem  the  trial  divisor  must 
always  be  written  in  the  form  x  —  a. 

Ex.  1 .    Factor  xi  +  6  x"'  +  3  x2  +  x  +  3. 

If  there  is  a  factor  of  the  form  x  —  a.  then  the  onty  possible  values 
of  a  are  the  various  divisors  of  '■>.  namely  +  1,  —  1,  +  ;>,  —  ;>- 

To  test  the  factor  x  +  1,  we  write  it  in  the  form  x  —  (—  1)  where 
a  —  —  I.     Substituting  —  1  for  x  in  the  polynomial,  we  have 

1-6  +  3-1  +  3  =  0. 

Hence  x  +  1  is  a  factor. 

On  substituting  +  1,  +  3,  —  3  for  x  successively,  no  one  reduces  the 
polynomial  to  zero.     Hence  x  —  1,  x  —  3,  x  +  3  are  not  factors. 

Ex.  2.     Factor  3  a?  -x2-A  x  +  2. 

If  x  —  a  is  a  factor,  then  a  must  be  a  factor  of  +  2.  We  therefore 
substitute,  +2,  —2.  +  1,  —1  and  find  the  expression  becomes  zero 
when  4-1  is  substituted  for  a;.  Hence  x  —  1  is  a  factor.  The  other 
factor  is  found  by  division  to  b<>  3  .'-  +  2x  —  2.  which  is  prime. 

1  [ence  3 ,3  _  X2  _  4  x  +  2  =  (x  -  1 ) (3  x-  +  2  x  -  2). 


THE  FACTOR    THEOREM  59 

EXERCISES 

Factor  by  means  of  the  factor  theorem : 

1.  3xi  —  2x2  +  5x  —  6.  6.    ra3  +  5m2  +  7ra  +  3. 

2.  2xs  +  3x2-3x-4.  7.   x4  +  3x5-3x2-7x  +  6. 

3.  2ar3  +  ar9-12a:  +  9.  8.    3rJ  +  5r2-7r-l. 

4.  a3 +  9  a-2 +  10 a: +  2.  9.    2  z3  +  7  z2  +  4z  +  3. 

5.  a3-3a  +  2.  10.    a3- 6 a2 +  11  a -6. 

11.  Show  by  the  factor  theorem  that  xk  —  ak  contains  the 
factor  x  —  a  if  A;  is  cm?/  integer. 

12.  Show  that  xk  —  ak  contains  the  factor  x  +  a  if  &  is  any 
even  integer. 

13.  Show  that  xk  +  ak  contains  the  factor  x-\-a  if  Tc  is  any 
ocW  integer. 

14.  Show  that  xk-\-ak  contains  neither  x-\-a  nor  x  —  a  as  a 
factor  if  k  is  an  even  integer. 

MISCELLANEOUS   EXERCISES   ON  FACTORING 

1.  20a3.i%-45a3.^3.  4.    16  x2  -  72  xy  +  81  y2. 

2.  24  am5ri2  -  375  am¥.  5.   162  a36  +  252  a262  +  98  a&3. 

3.  432  a)As  +  54  ars\  6.    48  afy  - 12  afy- 12  a;2?/ +3?/. 

7.  12  a2te2  +  8  a&V  +  18  a26.r?/  + 12  ab2xy. 

8.  18  x3#- 39  a;2?/2 +  18  ay3.         16.    as-f.  17.    a}6-y™. 

9.  4ar-9a,v/+6a;-9?/+4a;+6.    18.    a8  +  «Y  +  ?A 

10.  6a;2-13a-?/  +  6?/2-3.r+2y.  19.  as  +  a  —  2. 

11.  6  a4  +  15 .1/7/°  + 9  ?/4.  20.  cf  —  lSaY  +  y*. 

12.  16.«4  +  24a-7/2  +  8  7/4.  21.  aI6-6«y  +  f. 

13.  15  a?4  +  24  x2y2  +  9  //4.  22.  .ir3  +  4ar  +  2a  —  1. 

14.  afi  +  ?/6.  15.    al2  +  yV2.  23.  3  ar3  +  2  s:2  —  7  a;  +  2. 


60  FACTORING 

24.  a8-3ay  +  ?/8.  26.    a3  +  9  a2  +  16  a +  4. 

25.  a8  +  a2  +  a  +  l.  27.    2  z4  +  or3?/  +  2  a,-2?/2  +  a;?/3. 

28.  ?«5  +  m^i  -{-  m8a2  +  m2a?  4-  ??ia4  +  a5. 

29.  (x-2y-(y-z)s. 

30.  a6  +  bG  4-  2  a6(a4  —  a2b2  +  &4). 

31.  afy5  +  xly4z  +  O/"3^2  +  xhfz*  -j-  a*?/z4  +  2s. 

32.  8as  +  6ab(2a-3b)-27b3. 

33.  a  (jb3  +  f)-  ax (x2  -  y2)  -y\x  +  ?/). 

34.  a3  -  &3  +  3  b-c  -3bc2  +  c8. 

35.  a4  +  2  a86  —  2  ab2c  -  b2c2. 

36.  a4  +  2  a3&  +  a2&2  -  a'b2  -  2  a2&2c  -  &V. 

SOLUTION  OF  EQUATIONS  BY  FACTORING 

94.  Many  equations  of  higher  degree  than  the  first  may  be 
solved  by  factoring.     (See  §§  144-146,  E.  C.) 

Ex.  1.    Solve  2  ar3  -  x2  -  5  x  -  2  =  0.  (1) 

Factoring  the  left  member  of  the  equation,  we  have 

(ar-2)(ar+l)(2x+l)  =  0.  (2) 

A  value  of  x  which  makes  one  factor  zero  makes  the  whole  left 
member  zero  and  so  satisfies  the  equation.  Hence  x  =  2,  x  =  —  1, 
x  =  —  \  are  roots  of  the  equation. 

To  solve  an  equation  by  this  method  first  reduce  it  to  the  form 
-4=0,  and  then  factor  the  left  member.  Put  each  factor  equal  to 
zero  and  solve  for  x.  The  results  thus  obtained  are  roots  of  the 
original  equation. 

Ex.  2.   Solve  ar3  - 12  x2  =  12  -  35  x.  (1) 

Transposing  and  factoring,  (x  —  4)(x2  —  8  x  +  3)  =  0.  (2) 

I  Icnce  the  roots  of  (1)  are  the  roots  of  x  —  4  =  0  and  x2  —  Sx  +  3  =  0. 
From  x  —  4  =  0,  x  =  4.  The  quadratic  expression  x'2  —  8  x  +  3  can- 
not be  resolved  into  rational  factors.     See  §  155. 


COMMON  FACTORS  AND  MULTIPLES  61 

EXERCISES 

Solve  each  of  the  following  equations  by  factoring,  obtain- 
ing all  roots  which  can  be  found  by  means  of  rational  factors. 

1.  x3  +  3  x2  =  28  x.  6.  2x3  +  3x  =  9x2-14. 

2.  6xs  +  8x-\-5  =  19x2.  7.  5x3  +  x2-14x  +  8  =  0. 

3.  x4  +  12  x2  +  3  =  7x3  +  9x.  8.  2  x3  -f-af'  =  14  a;  — 3. 

4.  12x3  =  20x2  +  5x  +  6.  9.  12  x4  +  U  x3+l  =  3  .r2  +  4  x. 

5.  x3-4r  =  4.i-  +  5.  10.  x5-4x4-40x3+6-x=58x2. 

COMMON   FACTORS    AND    MULTIPLES 

95.  If  each  of  two  or  more  expressions  is  resolved  into 
prime  factors,  then  their  Highest  Common  Factor  (H.  C.  F.)  is  at 
once  evident  as  in  the  following  example.     See  §  182,  E.  C. 

Given  (1)         x4  -  y4  =  (x2  +  y*)(x  +  y)(x  -  y), 
(2)         x6  -if  =  (x3  +  >f)  (x3  -  ?/3) 

=  (x  +  y)(x2  -  xy  +  i/)(x  -  ij)(x2  +  xy  +  if). 
Then  (x  +  y)(x  —  y)  =  x2  -  >f  is  the  II.  C.  F.  of  (1)  and  (2). 

In  case  only  one  of  the  given  expressions  can  be  factored  by 
inspection,  it  is  usually  possible  to  select  those  of  its  factors, 
if  any,  which  will  divide  the  other  expressions  and  so  to  deter- 
mine the  H.  C.  F. 

Ex.    Find  the  H.  C.  F.  of  6  .f3  +  4  x2  -  3  x  -  2, 
and  2  x4  +  2  x3  +  x2  —  x  —  1. 

By  grouping  we  rind  : 

6  x3  +  4  x2  -  3  x  -  2  =  2  x2  (3  x  +  2)  -  (3  x  +  2) 
=  (2x2-  l)(3x  +  2). 

The  other  expression  cannot  readily  be  factored  by  any  of  the 
methods  thus  far  studied.  However,  if  there  is  a  common  factor,  it 
must  be  either  2  x2  —  1  or  3  x  +  2.  We  see  at  once  that  it  cannot  be 
3  x  +  2.  (Why  ?)  By  actual  division  2  x2  —  1  is  found  to  be  a  factor 
of  2  x4  +  2  x3  +  x2  -  x  -  1.     Hence  2  x2  -  1  is  the  H.  C.  F. 


62  FACTORING 

96.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
expressions  is  readily  found  if  these  are  resolved  into  prime 
factors.     See  §  185,  E.  C. 

Ex.  1.    Given     6  abx  -  6  aby  =  2-3  ab(x  -  y),  (1) 

SaPx  +  Saiy  =  2sa2(x  +  IJ),  (2) 

3G  b%x?  -  y*)(x  +  y)  =  2-&V(x  -y){x  +  y)\  (3) 

The  L.  CM.  is  23  ■  S2a2bs  (x  -  y)(x  +  y)2,  since  this  contains  all 
the  factors  of  (1),  all  the  factors  of  (2)  not  found  in  (1),  and  all 
the  factors  of  (3)  not  found  in  (1)  and  {'2),  with  no  factors  to  spare. 

In  case  only  one  of  the  given  expressions  can  be  factored  by 
inspection,  it  may  be  found  by  actual  division  Avhether  or  not 
any  of  these  factors  will  divide  the  other  expressions. 

Ex.  2.    Eind  the  L.  C.  M.  of  6  x*  -  x2  +  4  x  +  3,  (1) 

and  6  0s  +  3  x2  -  10  x  -  5.  (2) 

(1)  is  not  readily  factored.  Grouping  by  twos,  the  factors  of  (2) 
are  3  x2  —  5  and  2  x  +  1.  Now  '■'> ./-  —  5  is  not  a  factor  of  (1).  (Why  ?) 
Dividing  (1)  by  2  x  +  1  the  quotient  is  3  x2  —  2  x  +  3. 

Hence       6  x3  -  x2  +  4  x  +  3  =  (2  x  +  1)  (3  x2  -  2  x-  +  3), 
6  x3  +  3  x2  -  10  x  -  5  =  (2  x  +  1)  (3  x2  -  5). 

Hence  the  L.  C.  M.  is  (2  x  +  1)  (3  x2  -  2  x  +  3)  (3  x2  -  5). 

Ex.  3.   Find  the  L.  C.  M.  of  a?  +  2 a- -a- 2,  (1) 

and  l0a3-3«2  +  4a  +  l.  (2) 

By  means  of  the  factor  theorem,  a  —  1,  a  +  1,  and  a  +  2  are  found 
to  be  factors  of  (1),  but  none  of  the  numbers,  1,  —  1,  —  2.  when  sub- 
stituted for  a  in  ("J)  will  reduce  it  to  zero.  Hence  (1)  and  (2)  have 
no  factors  in  common.  The  L.  CM.  is  therefore  the  product  of  the 
two  expressions  ;  viz.  (a  +  \)(<t  -  l)(a  +  2)(10«3  -  3a2  +  4  a  +1). 

97.  The  H.C.  F.  of  three  expressions  may  be  obtained  by 
finding  the  H.  C.  F.  of  two,  and  then  the  H.  C.  F.  of  this  result 
and  the  third  expression.  Similarly  the  L.C.M.  of  three  expres- 
sions may  be  obtained  by  finding  the  L.  C.  M.  of  two  of  them, 
and  then  the  L.  C.  M.  of  this  result  and  the  third  expression. 

This  may  be  extended  to  any  number  of  expressions. 


COMMON  FACTORS  AND   MULTIPLES  63 

EXERCISES 

Find  the  H.  C.  F.  and  also  the  L.  C.  M.  in  each  of  the 
following : 

1.  x'2  4-  y2,  x6  +  ?/6.  2.  x-  +  xy  4-  y2,  x3  -  y3. 

3.  x2  —  5x  —  6,   a;2 -2.x- -3,   a;2  4- 19  .»  4- 18. 

4.  x4  —  6.r24-l,   ar34-ar  — 3.c  +  l,    £34-3ar4-#  — 1. 

5.  162a%  +  252o262  +  9a63,   54a3  +  42a26. 

6.  2.r3  +  x2-8a-  +  3,    a2  +  2a;-l. 

7.  3r3  +  5?-2-7r-l,   3r2+8r  +  l. 

8.  a3  — 3  a2 +  4,    ax  —  ab  —  2x  +  2b. 

9.  a6  +  2  a363  +  66  -  2  a46  -  2  a&4,   a3  -2ab  +  b3. 

10.  8  a3  -  36  a26  +  54  c^2  -  27  63,   4a2-9  62. 

11.  36  o4- 9  a2 -24  a -16,   12  «3- 6  a2 -8  a. 

12.  2  if  +  4%  +  3cy  +  6  be,   y2- 3  by -10  b2 

13.  cc16  — 2/1G,   a-8  —  ?/8,    x4  —  y\ 

14.  m3  4-  8  m2  +  7  m,    m3  4-  3  m2  —  m  —  3,   m3  —  7m—  6. 

98.  An  important  principle  relating  to  common  factors  is 
illustrated  by  the  following  example: 

Given  x2  +  7x  +  10  =  (x  +  5)(.x  +  2),  (1) 

and  a;2  —  x  —  6  =  (ar  —  3)  (a;  +  2).  (2) 

Add  (1)  and  (2),  2 x2  +  6 x  +  4  =  2(x  +  l)(.r  +  2).  (3) 

Subtract  (2)  from  (1),  8x  +  1G  =  S(x  +  2).  (4) 

We  observe  that  x  +  2,  which  is  a  common  factor  of  (1)  and  (2), 
is  also  a  factor  of  their  sum  (3),  and  of  their  difference  (4).  This 
example  is  a  special  case  of  the  following : 


64  FACTORING 

99.  Theorem  1.  A  common  factor  of  two  expressions  is 
also  a  factor  of  the  sum  or  difference  of  any  multiples  of 
those  expressions. 

Proof.     Let  A  and  B  be  any  two  expressions  having  the  common 

factor  f.     Then  if  k  and  I  are  the  remaining  factors  of  A   and  B 

respectively, 

A  =fk  and  B=  ft. 

Also  let  mA  and  nB  be  any  multiples  of  A  and  B. 

Then  mA  =  mfk  and  nB  —  nfl,  from  which  we  have : 

mA  ±  ?i B  =  mfk  ±  nfl  =  f(mh  ±  id). 

Hence  f  is  a  factor  of  mA  ±  nB. 

100.  Theorem  2.  //  f  is  a  factor  of  mA  ±  nB  and  also  of 
A.  then  f  is  a  factor  of  B,  provided  n  has  no  factor  in  com- 
mon with  A. 

Proof.  Let  /  be  a  factor  of  n>A  ±  nB  and  also  of  A,  where  mA 
and  nB  are  integral  multiples  of  the  expressions  A  and  B. 

Then   mA  ±  nB,  — ,  and  —  may  each  be  reduced  to  an  integral 

expression  by  cancellation. 

Now  mA±nB  =  ?lA±ilB.    SinCe  2^  is  integral,  it  follows 

nB       f  f  f  f 

that  —  is  also  integral.     That  is,  /  is  a  factor  of  nB.     But  / 

is  not  a  factor  of  n  since  it  is  a  factor  of  A,  and  by  hypothesis 
n  and  A  have  no  factor  in  common.     Hence  /  is  a  factor  of  B. 

101.  By  successive  applications  of  the  above  theorems  it  is 
possible  to  find  the  H.  C.  F.  of  any  two  integral  expressions. 

Ex.  1.   Find  the  H.  C.  F.  of  9  x*  -  x-  +  2  x  - 1,  (1) 

and  27cc5  +  8<B2-3a;  +  l.'  (2) 

Multiplying  (1)  by  Zx  and  subtracting  from  (2)  we  have 
•_'7  xB  +  8  x2  -  3  x  +  1 
27  x5  -  3  x3  +  6  x2  -  3  x 

3x3  +  '2x2  +  l  (3) 

By  theorem  1,  any  common  factor  of  (1)  and  (2)  is  a  factor  of  (3). 


COMMON  FACTORS  AND  MULTIPLES  65 

Calling  expressions  (1)  and  (2)  B  and  A  respectively  of  theorem  2, 
then  (3)  is  A  —  3  x  •  B ;  and  since  the  multiplier,  3  x,  has  no  factor  in 
common  with  (2),  it  follows  from  the  theorem  that  any  common 
factor  of  (3)  and  (2)  is  a  factor  of  (1),  and  also  that  any  common 
factor  of  (3)  and  (1)  is  a  factor  of  (2).  Hence  (1)  and  (3)  have  the 
same  common  factors,  that  is,  the  same  II.  C.  F.  as  (1)  and  (2).  There- 
fore we  proceed  to  obtain  the  H.  C.  F.  of 

9/4-r+2z-l,  (1) 

and  3x3  +  2a»+l.  (3) 

Multiplying  (3)  by  3  a;  and  subtracting  from  (1)  we  have 

—  6  x3  —  x2  —  x  —  1.  (4) 

By  argument  similar  to  that  above,  (3)  and  (4)  have  the  same 
H.  C.  F.  as  (1)  and  (3)  and  hence  the  same  as  (1)  and  (2).  Multi- 
plying (3)  by  2  and  adding  to  (4)  we  have, 

3x2-x  +  l.  (5) 

Then  the  H.  C.  F.  of  (5)  and  (3)  is  the  same  as  that  of  (1)  and  (2). 
Multiplying  (5)  by  x  and  subtracting  from  (3),  we  have 

3  x2  -  x  +  1 .  (6) 

Then  the  H.  C.F.  of  (5)  and  (0)  is  the  same  as  that  of  (1)  and  (2). 
But  (5)  and  (6)  are  identical,  that  is,  their  H.  C.  F.  is  ox2  —  x  +  1. 
Hence  this  is  the  H.  C.  F.  of  (1)  and  (2). 

The  work  may  be  conveniently  arranged  thus  : 


(1)     9  x4              -    x2  +  2  x  -  1 

27  x5 

+  8  x2  -  3  x  +  1 

(?) 

J)  x4  +  6  x3            +  3  x 

27  x5- 

-  3  x3  +  6  x2  —  3  x 

(4)            -  6  x3  -    x2  -    x  -  1 

3  x3  +  2  x2           +1 

(3) 

6i3+4x2           +2 

3  x3  —    x2  +    x 

(5)  3  x2  -x  +  1  3  x-  -    x  +  1     (G ) 

The  object  at  each  step  is  to  obtain  a  new  expression  of  as  low  a 
degree  as  possible.  For  this  purpose  the  highest  powers  are  elimi- 
nated step  by  step  by  the  method  of  addition  or  subtraction. 

E.g.  In  Ex.  1,  x5  was  eliminated  first,  then  x4,  and  then  x3. 

By  theorems  1  and  2,  each  new  expression  contains  all  the  factors 
common  to  the  given  expressions.  Hence,  whenever  an  expression  is 
reached  which  is  identical  with  the  preceding  one,  this  is  the  H.  C.F. 


6G  FACT  OBI  Mi 

102.    The  process  is  further  illustrated  as  follows: 

Ex.  2.    Find  the  H. C.  F.  of  2 ar*  -2a?-3x-2, 
and  3x3  —  x2  —  2x—16. 

Arranging  the  work  as  in  Ex.  1,  we  have 

3x3-      x2-    2  x  -    16     (2) 
6x3-    2x2-    4x-    32 
6x3-    6a:2-    9x-      6 


(1) 

2x3-2x2-    3x-2 

4  x3  —  4  x*2  —    6  x  —  4 

4  a;3  +  5  x2  _  oe  x 

(1) 

-  9  x2  +  20  x  -  4 

9x2-18a: 

(7) 

2x-4 

(8) 

x-2 

4X2+    5a._    26     (3) 
36  a-2  +  45  x  -  234 
-36x2  +  80x-    16 

125  x  -250     (5) 

x  -      2     (6) 


Explanation.  To  eliminate  x3,  we  multiply  (1)  by  3  and  (2)  by  2 
and  subtract,  obtaining  ('■)). 

To  eliminate  x2  from  (3),  we  need  another  expression  of  the  second 
degree.  To  obtain  this  we  multiply  (1)  by  2  and  (;,>)  by  x  and  sub- 
tract, obtaining  (1). 

Using  (4)  and  (3),  we  eliminate  x'-\  obtaining  (5).  Since  (.1)  con- 
tains all  factors  common  to  (1)  and  (2),  and  since  125  is  not  such  a 
factor,  this  is  discai'ded  without  affecting  the  H.  C.  F..  giving  (6). 

Multiplying  (6)  by  9  and  adding  to  (4)  we  have  (7).  Discarding 
the  factor  2  gives  (8)  which  is  identical  with  (6).  Hence  x  —  2  is  the 
H.  C.  F.  sought. 

103.  Any  monomial  factors  should  be  removed  from  each 
expression  at  the  outset.  If  there  are  such  factors  common  to 
the  given  expressions,  these  form  a  part  of  the  H.  C.  F. 

When  this  is  done,  then  any  monomial  factor  of  any  one  of 
the  derived  expressions  may  be  at  once  discarded  without 
affecting  the  H.  C.  F..  as  in  (5)  of  the  preceding  example. 

Tti  this  way  a  bo  the  hypothesis  of  theorem  2  is  always  fulfilled; 
namely,  thai  at  every  step  the  multiplier  of  one  expression  shall  have 
no  factor  in  common  with  the  other  expression. 


COMMON  FACTORS  AND  MULTIPLES  (57 

Ex.  3.   Find  the  H.  C.  F.  of  3  Xs  -  7  x2  4-  3  x  -  2, 

and  x4  —  x3  —  x2  —  x  —  2. 

(1)  3x3-    7x2  +    3x-    2  xi  -     xs-     x2  -     x  -  2  (2) 

12  x3  -  28  x2  +  12  x  -    8  3  x4  -  3  x3  -  3  a;2  -  3  x  -  6 

12  x3  -  18  a:2  -    3  x  -  18  3  a,4  -  7  x3  4-  3  .r2  -2  a: 


(1)  -  10x2  +  15 x  +  10  4x3-0x2-    x-6    (3) 

(5)       -5(2x+l)(x-2). 

Explanation.  To  eliminate  x4,  we  multiply  (1)  by  x  and  (2)  by  3 
and  subtract,  obtaining  (3). 

To  eliminate  x3,  we  multiply  (1)  by  4  and  (3)  by  3  and  subtract, 
obtaining  (1). 

At  this  point  the  work  may  be  shortened  by  factoring  (4)  as  in 
(5).  We  may  now  reject,  not  only  the  factor  —  5,  but  also  2x  +  1, 
which  is  a  factor  of  neither  (1)  nor  (2),  since  2x  does  not  divide  the 
highest  power  of  either  expression.  But  x  —  2  is  seen  to  be  a  factor 
of  (2),  by  §§  91,  92,  and  hence  it  is  a  common  factor  of  (2)  and  (4) 
aud  therefore  of  (1)  and  (2).     Hence  x  —  2  is  the  H.  C.  F.  sought. 

EXERCISES 

Find  the  H.C.F.  of  the  following  pairs  of  expressions: 

1.  a?  +  6  a2  +  6  a  +  5,  a3  +  4  a2  -  4  a  +  5. 

2.  xi-2x*-2xi  +  5x-2,  xi-4x3  +  6x2-5x  +  2. 

3.  2  a3 -9. ^-13 z-4,  x3 -  12 x2  +  31  x  +  28. 

4.  x*  _  5  rf  _|_  3  x  -  2,  x*  -  3  x'  +  3  x2  -  3  x  +  2. 

5.  2  Xs  -  9  x2  +  8  a-  -  2,  2  x3  +  5  or  —  5  x  +  1. 

6.  3  a4  -  2  a3  + 10  a2  -  6  a  +  3,  2  a4  4-  3  a3  +  5  a2  +  9  a  -  3. 

7.  15  .r4  + 19  a3 -44  a;2 -15  A- +  9, 

15  x*  -  6  a;3  +  51.»2  + 11  x  -  15. 

8.  rs  +  2^-2r8-8r2-7r-2,  ^-2r4-  2r»+  4»-2+  r-2. 

104.  The  following  theorem  enables  us  to  find  the  L.  C.  M.  of 
two  expressions  by  means  of  the  method  which  has  just  been 
used  for  finding  the  H.  C.  F. 


68  FACTORING 

Theorem  3.  The  L.C..M.  of  two  expressions  is  equal  to 
the  product  of  either  expression  and  the  quotient  obtained 
by  dividing  the  other  by  the  II.  C.  F.  of  the  two  expressions. 

Proof.  Let  A  and  B  be  two  expressions  whose  H.C.  F.  is  F  so  that 
A  =mF and  B=  nF.  Hence  the  L.  CM.  of  A  and  B  is  mnF.  But 
mnF  =  mnF =  mB.  Also  mnF  =  nmF  =  nA.  Therefore  the  L.  CM. 
is  cither  mB  or  nA,  where  m  =  A  h-  F  and  n  —  B  -=-  F. 

Ex.   Find  the  L.  C.  M.  of  9  *4  -  a-2  +  2  x  -  1,  (1) 

and  27  x5  4-  8  x-  —  3  »  +  1.  (2) 

The  II.  CF.  was  found  in  §  101  to  be  3  a;2  -x  +  1. 
Dividing  (1)  by  3  x2  —  x  +  1  we  have  3  x'2  +  x  —  1. 

Hence  the  L.  C  M.  of  (1)  and  (2)  is 

(27  x5  +  8  x2  -  3  x  +  1)  (3  x2  +  x  -  1) . 

EXERCISES 

Find  the  L.  C.  M.  of  each  of  the  following  sets. 

1.  a4  4-  a?  +  2  a2  —  a  +  3,  a4 +  2  a3 +  2  a2  — a +  4. 

2.  a3-  6a2  4-  11  a  -  6,  a3-  9  a2  +  20  a  -  24. 

3.  2  a3+3  a26  -  2  a&2  -  3  &33  2  a4  -  a86  -  2  a2&2  4-  4  a&3-  3  64. 

4.  2a8-a2&-13a&2-668, 

2  a4-  5  a3&  -  11  aVr  4-  20  oi>3  +  12  b\ 

5.  4  as_  15  a2_  5  a  _  3>  8  a4_  34  as_j_  5  a->  _  a  +  3, 
2a8-7a2+lla-4. 

6.  a*  +  a2  +  1,  a3  +  2  a2  -  2  a  +  3. 

7.  2&8-&2Z-13fcZ24-5Zs,  3fc3-l(>fr"-7  +  24A72-7  73. 

8.  12  v4-  20  r\s  -  15  rs-+  35  rs8-  12  s4, 
6r8-7r2s-llrs2  +  12s8. 

9.  2  tr-7 a2+6a-2,  a3+2 a2-  13 a+10,  a3+  0 a2 4-6 a+5. 

10.    x"3—  xi f+  yx-  —  if,  2  x8  4-  .r//  4-  •>'!/-+  -  ."''• 
2.  r!+;  i.r-y  +3  a*?/2  +2^. 


CHAPTER   VI 
POWERS  AND  ROOTS 

105.  Each  of  the  operations  thus  far  studied  leads  to  a  single 
result. 

E.g.  Two  numbers  have  one  and  only  one  sum,  §  2,  and  one  and 
only  one  product,  §  7. 

When  a  number  is  subtracted  from  a  given  number,  there  is  one 
and  only  one  remainder,  §  6. 

When  a  number  is  divided  by  a  given  number,  there  is  one  and 
only  one  quotient,  §  11- 

We  are  now  to  study  an  operation  which  leads  to  more  than 
one  result ;  namely,  the  operation  of  finding  roots. 

Thus  both  3  and  —  3  are  square  roots  of  9,  since  3-3  =  9,  and  also 
(_  3)(_  3)  =  9;  this  is  often  indicated  by  V9  =  ±  3.     See  §  114. 

106.  The  operations  of  addition,  subtraction,  multiplication, 
and  division  are  possible  in  all  cases  except  dividing  by  zero, 
which  is  explicitly  ruled  out,  §§  24,  25. 

Division  is  possible  in  general  because  fractions  are  admitted  to 
the  number  system,  and  subtraction  is  possible  in  general  because 
negative  numbers  are  admitted.     Thus  7  h-  3  =  2J,  5  —  7  ~  —  2. 

107.  The  operation  of  finding  roots  is  not  possible  in  all 
cases,  unless  other  numbers  besides  positive  and  negative  in- 
tegers and  fractions  are  admitted  to  the  number  system. 

E.g.    The  number  V2  is  not  an  integer  since  l2  =  1  and  22  =  4. 
Suppose  V2  =  -  a  fraction  reduced  to  its  lowest  terms,  so  that  a 

and  b  have  no  common  factor.     Then  —  =  2.     But  this  is  impossible, 

b~ 

for  if  b'1  exactly  divides  a2,  then  a  and  b  must  have  factors  in  com- 
mon.    Hence  V2  is  not  &  fraction. 

69 


70  POWERS    AND    ROOTS 

108.  If  a  positive  number  is  not  the  square  of  an  integer  or 
a  fraction,  a  number  may  be  found  in  terms  of  integers  and 
fractions  whose  square  differs  from  the  given  number  by  as 
little  as  we  please.     See  p.  228,  E.  C. 

E.g.  1.41,  1.414,  1.4141  are  successive  numbers  whose  squares  differ 
by  less  and  less  from  2.  In  fact  (1.4141)-  differs  from  2  by  less  than 
.0004,  and  by  continuing  the  process  by  which  these  numbers  are 
found,  §  170,  E.  C,  a  number  may  be  reached  whose  square  differs 
from  2  by  as  little  as  we  please. 

1.41,  1.414,  1.4141,  etc.,  are  successive  approximations  to  the 
number  which  we  call  the  square  root  of  2,  and  which  we  represent  by 
the  symbol,  V2. 

109.  Definition.  If  a  number  is  not  the  A'th  power  of  an 
integer  or  a  fraction,  but  if  its  feth  root  can  be  approximated 
by  means  of  integers  and  fractions  to  any  specified  degree  of 
accuracy,  then  such  a  kth  root  is  called  an  irrational  number. 
See  §  36. 

E.g.  V2,  V2,  a/5,  etc.,  are  irrational  numbers,  whereas  V4,  V8, 
are  rational  numbers. 

It  is  shown  in  higher  algebra  that  irrational  numbers  corre- 
spond to  definite  points  on  the  line  of  the  number  scale,  §  40, 
E.  C,  just  as  integers  and  fractions  do. 

We,  therefore,  now  enlarge  the  number  system  to  include 
irrational  numbers  as  well  as  integers  and  fractions. 

It  will  be  found  also  in  higher  work  that  there  are  other  kinds  of 
irrational  numbers  besides  those  here  defined. 

The  set  of  numbers  consisting  of  all  rational  and  irrational 
numbers  is  called  the  real  number  system. 

110.  Even  with  the  number  system  as  thus  enlarged,  it  is 
still  no1  possible  to  find  roots  in  all  cases.  The  exception  is 
the  even  root  of  a  negative  number. 


THE  COMPLEX  NUMBER  71 

E.g.  V  —  4  is  neither  +  2  nor  -  2,  since  (+  2)'2  =  +  4  and  (-  2)'2 
=  +  4,  and  no  approximation  to  this  root  can  be  found  as  in  the  case 

of  V2. 

111.  Definition.  The  indicated  even  root  of  a  negative  num- 
ber, or  any  expression  containing  such  a  root,  is  called  an 
imaginary  number,  or  more  properly,  a  complex  number.  All 
other  numbers  are  called  real  numbers. 

E.g.  V—  4,  V—  2,  1  +  V  —  2,  are  complex  numbers,  while  5,  a/2, 
1  +  V2  are  real  numbers. 

Complex  numbers  cannot  be  pictured  on  the  line  which  represents 
real  numbers,  but  another  kind  of  graphic  representation  of  complex 
numbers  is  made  in  higher  algebraic  work,  and  such  numbers  form 
the  basis  of  some  of  the  most  important  investigations  in  advanced 
mathematics. 

112.  With  the  number  system  thus  enlarged,  by  the  admis- 
sion of  irrational  and  complex  numbers,  we  have  the  following 
fundamental  definition. 

(-s/7iy  =  n. 

That  is,  a  £th  root  of  any  number  n  is  such  a  number  that, 
if  it  be  raised  to  the  kt\\  power,  the  result  is  n  itself. 

E.g.    (aV2)3  =  2,   (VI)2  =  4,  (V^  =  _2. 

The   imaginary  or  complex  unit  is  V— 1.     By  the  above 

definition  we  have  

(V-l)2  =  -l. 

In  operating  upon  complex  numbers,  they  should  first  be  ex- 
pressed in  terms  of  the  imaginary  unit. 

E.g.    V^2  =  V2  ■  V^l,  V-  1(3  =  VT6  •  V^l  =  4V^T. 

V^I.  v/3q=  (VI-  V^T)(\/U  V~l)=2.3(V^T)2=-<>. 

\/^4  +  \/^9=  VI-  V^l+  VU  •  \/3I  =  (2  +  3)V^T=5\/^l. 

V^i6  =  Vie-  V3i^vi6  =  4 

V^  "  V9  •  V^l        V9      3 


72  POWERS  AND   ROOTS 

113.  By  means  of  irrational  and  complex  numbers  it  can  be 
shown  that  every  number  has  two  square  roots,  three  cube 
roots,  four  fourth  roots,  etc.     See  §  195,  Ex.  17-20. 

E.g.  The  square  roots  of  9  are  +  3  and  —  3.  The  square  roots 
of  -  9  are  ±  V^9  =  ±  3  V~l.  The  cube  roots  of  8  are  2,  - 1  +  V^3 
and  —  1  —  V—  3.  The  fourth  roots  of  16  are  +2,  —  2,  +  2V—  1 
and  -2V3T. 

Any  positive  real  number  has  two  real  roots  of  even  degree, 
one  positive  and  one  negative. 

E.g.    Vlti  =  ±  2.    The  square  roots  of  3  are  ±  V3. 

Any  real  number,  positive  or  negative,  has  one  real  root  of 
odd  degree,  whose  sign  is  the  same  as  that  of  the  number  itself. 

E.g.    ^27  =  3  and  v^-32  =  -  2. 

114.  The  positive  even  root  of  a  positive  real  number,  or  the 
real  odd  root  of  any  real  number,  is  called  the  principal  root. 

The  positive  square  root  of  a  negative  real  number  is  also 
sometimes  called  the  principal  imaginary  root. 

E.g.  2  is  the  principal  square  root  of  4,  3  is  the  principal  4th  root 
of  81 ;  —  4  is  the  principal  cube  root  of  —  01 ;  and  +  V  —  3  is  the 
principal  square  root  of  —  3. 

Unless  otherwise  stated  the  radical  sign  is  understood  to  in- 
dicate the  "principal  root. 

The  only  exception  in  this  book  is  in  such  cases  as,  v'4  =  ±  2. 
where  it  represents  either  square  rout,  But  in  such  expressions  as 
1  +  V2,  3  ±  V6,  etc.,  the  principal  root  only  is  understood. 

In  all  cases  it  is  easily  seen  from  the  context  in  what  sense  the  sign 
is  used. 

When  it  is  desired  to  designate  in  particular  the  principal 
root,  the  symbol  V        is  used. 

E.g.  VlG'  =  2,  while  VlG  might  stand  indifferently  for  2.  —2. 
2/-  1",  or  _2V^1. 

\  8  —  2.  while  \  s  might  represent  2,  —  1  +  V—  3,  or  —  1  —  V—  3. 


THEOREMS    ON  POWERS  ANU   ROOTS  73 

THEOREMS   ON   POWERS   AND   ROOTS 

115.  Theorem  1.  The  nth  power  of  the  kth  power  of  any 
base  is  the  nktli  power  of  that  base. 

Proof.    Let  n  and  k  be  any  positive  integers  and  let  b  be  any  base. 

Then  (&*)»  =  bk  ■  bk  .  bk  ...  to  n  factors.  §  124,  E.  C. 

—  0k+k+k...  to  n  terms  _  Qnk^  g  4.3 

Hence  (b")a  =  b'"c. 

Corollary  (bk)n  =  (bn) k  =  b"k. 

E.g.    (23)2  =  (2-)3  =  26  =  64. 

116.  Theorem  2.  The  nth  power  of  the  product  of  several 
factors  is  the  product  of  the  nth  powers  of  those  factors. 

Proof.    Let  k,  r,  and  n  be  any  positive  integers.     Then 

'akb')n  =  (akb'-)  •  (ahbr)  •  •  •  to  n  factors,  §  124,  E.C. 

=  (a*  •  a*  . . .  to  n  factors)  (6*  •  bk  •  ■  ■  to  n  factors)  §§  8,  9 

=  (a*)» -(brY,  §124,  E.C. 

Hence,  (akb'J'=  ankbnr.  §115 

E.g.(;23-3-2)*  =  2r>-3i. 

117.  Theorem  3.  The  nth  power  of  the  quotient  of  two 
numbers  equals  the  quotisnt  of  the  nth  powers  of  those 
mi  i  nbers. 

Proof.    We  have  (£- )    =  £-  •  ~  •  ? to  n  factors  §  124,  E.  C. 

\br  I  If      It'       b' 

ak  •  ak  ■  ak  •  •  •  to  n  factors  c  ,  no   ^,  r, 

= — ; —    — .  §  lyo,  rj.L. 

br  ■  b1'  ■  b1'  •  •  •  to  ti  factors 

(ak\n       an» 

Hence'  [v)  =f^-  §115 

„  /28\2      26      64 

E-g-        y  =3i=8i- 


74  POWERS   AXD   ROOTS 

118.  It  follows  from  theorems  1,  2,  and  3  that: 

Any  positive  integral  power  of  a  monomial  is  found  by 
multiplying  the  exponents  of  the  factors  by  the  exponent  of  the 
power. 

119.  Theorem  4.  The  principal  rtli  root  of  the  krth 
power  of  any  positive  real  number  is  a  power  of  that 
number  ivhose  exponent  is  kr^-r  =  k. 

Proof.  Let  k  and  r  be  positive  integers  and  let  b  be  any  positive 
real  number. 

We  are  to  prove  that  Vb1^  =  hk. 

From  theorem  1,  (''*)''  =  &**"• 

Hence  by  definition  bk  is  an  rtli  root  of  ?/''.  and  since  bk  is  real  and 
positive,  it  is  the  principal  rth  root  of  t,'-r  (§  114). 

That  is,  \/W  =  bkr^r  =  &*. 

E.g.    VF  =  3^-2  =  32  =  9  ;  y/2^  =  212*4  =  23  =  8. 
But  it  does  not  follow  that 

^/(3oyn?  =  (_  2)U+4  =  (_  o)3  =  _  8) 

since  (-  2)12  =  (2)12  and  hence  ^/(-2)12'  =  ^2^  =  +  8. 

The  corresponding  theorem  holds  when  6  is  negative  if  r  is 
odd  and  also  when  b  is  negative  if  fc  is  even. 

E.#.    ^(-2)«'  =  (-2)6-3=(-2)2=4;  v^(-  2)ls'  =(-  2)s  =  -32. 

120.  Theorem  5.  The  principal  rtli  root  of  the  product 
of  two  positive  real  numbers  equals  the  product  of  the 
principal  rth  roots  of  the  number. 

Proof.     Let  a  and  b  be  any  positive  real  numbers  and  lei  r  be  any 
positive  integer. 
We  are  to  prove  v«&  ■=  Va  •  Vb  • 

We  have  ({/?  •  ^)r  =  ( Va  )r  •  V5" )'"  §  1 16 

=  a-&.  §112 

Hence,  a&  =  (-v^u"  •  v^  )r.  §  3 


THEOREMS   ON  POWERS  AND   ROOTS 


75 


Taking  the  principal  rth  root  of  both  members, 
we  have  VctF  —  Va '  •  VV . 

When  r  is  even  the  corresponding  theorem  does  not  hold 
if  a  and  b  are  both  negative. 


9'. 


For  example,  it  is  not  true  that  \/(—  ■!)(—  !>)'  =  V  —  -i 
For    V(-4)(-  9)  =  VW  =  6  ;  while    V-  4'  •  V^1 

=  2V^T'.3    V^l'=  6(V3T)2=_6.      See  §  112 

121.  Theorem  6.  The  principal  rth  root  of  the  quotient 
of  two  positive  real  numbers  equals  the  quotient  of  the 
principal  rth  roots  of  the  numbers. 

Proof.     Let  a  and  b  be  any  positive  real  numbers  and  let  r  be  any 
positive  integer. 
w  J-  ''/"        ^« 

We  are  to  prove  \h=  —  • 

™    vs 

Va 


We  have 


(gV=&2Za  §§  117,  112 


Hence,  taking  the  principal  ?-th  root  of  both  members, 
we  have 


r\a  _  Va 


E.g.    J^=^K  =  ±;  <B,=^8:=^2  =  _2. 
V25      v^     5  '    >  27         </27         3  3 

The  corresponding  theorem  does  not  hold   when  r  is  even 
if  a  is  positive  and  b  is  negative.     Thus  it  is  not  true  that 


x 


4 

-  9 


VI1 


3V^T     3(V"^T)2 


2v^T  2     

:~ r=-3^-i'- 


But  we  have 


v; 


i=%(-»=i 


If  r  is  odd,  the  theorem  holds  for  all  real  values  of  a  and  b. 


76  POWERS  AND  ROOTS 

122.    From  theorems  4,  5,  6,  it  follows  that : 

If  a  monomial  is  a  perfect  power  of  the  kth  degree,  its  kth 
root  may  be  found  by  dividing  the  exponent  of  each  factor  by  the 
index  of  the  root. 

In  applying  the  above  theorems  to  the  reduction  of  algebraic 
expressions  containing  letters,  it  is  assumed  that  the  values  of  the 
letters  are  such  that  the  theorems  apply. 

EXERCISES 

Find  the  following  indicated  powers  and  roots,  and  reduce 
each  expression  to  its  simplest  form : 

1.  (crW)7.  4.  (ax-yY+xs+^.  7.  (3"'  •  4' •  2y-\ 


2.  (2a+b  •  3°  •  5b)a-h.       5.    (.<ii/V+y)-r--v.  8.   "v  31'"  •  2a  •  530' 


a'h'rr'     Y  „     f?rb2>rin\-  Q        3  —  2< 


V  3TbcA  J  *     04rV/i;" 

10.  (a»4»-i6TO-»c-)»+  13    -^3.^  .  4._» .  #?-*• 

11.  (3«+*  •  4B"7  •  6-1)*.  14.    V64T25.256-6251. 

12.  2V3fia  •  4-a  •  58a  •  74a'.  15.    v  27'- 125-64.3". 

16.    (a  -  b)m~n(b  -  c)m~n(a  +  6)™-". 


17        /(a  -  6)8(ag  +  2  a&  +  b3) 
*         (a  -  &)4(a  +  6)2 

/(4x-2  +  4a;  +  l)(4x2-4.r  +  l)' 
'v  36  x*  -  12  a2  +  1 


19.    v(-  343)(-  27)./ (a  +  6)3«  . 

20      3/(-  8)(-  27)(-  125)aW 
\(_  1)(_  512)(1000)a;15ai/21 


ROOTS   OF  POLYNOMIALS  77 


ROOTS  OF  POLYNOMIALS 


123.  In  the  Elementary  Course,  pp.  221-224,  it  was  shown  that 
the  process  for  finding  the  square  root  of  a  polynomial  is  obtained 
by  studying  the  relation  of  the  square,  a?  +  2  ab  +  b2,  to  its 
square  root,  a  +  b. 

In  like  manner  the  process  for  finding  the  cube  root  of  a 
polynomial  is  obtained  by  studying  the  relation  of  the  cube, 
a?  +  3  d2b  +  3  ab'2  +  b:i  or  a8  -f-  6(3  a2  +  3  ab  +  ft2),  to  its  cube 
root,  a  +  b. 

An  example  will  illustrate  the  process. 

Ex.  1.    Eind  the  cube  root  of 

27  m3+ 108  m?n  + 144  mn2+  64  n3. 

Given  cube,  27  m3+108  »r2/i+144  mn2+64  n3 13  m+4  n,  cube  root 

a3  =  27  »i3  1st  partial  product 


3  a*  =  27  m2 
3a6  =  36  inn 
62  =  16  n2 
3  a2+3  a6+62=27  m2+36  mn+16  ?i2 


L08  /c-« M44  //m2+(i4  n3,         1st  remainder 
108  m2n-|-144  mn2+64  ??3=  6(3  «2+3  ab+b*) 


Explanation.  The  cube  root  of  the  first  term,  namely  om,  is  the 
first  term  of  the  root  and  corresponds  to  a  of  the  formula.  Cubing 
3  m  gives  27  in3  which  is  the  a3  of  the  formula. 

Subtracting  27  m3  leaves  108  m'2n  +  111  mri2  +  61  n3,  which  is  the 
6(3  a2  +  3  rtft  +  62)  of  the  formula. 

Since  b  is  not  yet  known,  we  cannot  find  completely  either  factor 
of  6(3  a2  +  3  ab  +  6'2),  but  since  a  has  been  found,  we  can  get  the  first 
term  of  the  factor  3  a'2 +  3  ab-\-b2 ;  viz.  3  rt"2  or  3(3  m)2  =  27  ?»'2,  which 
is  the  partial  divisor.  Dividing  108  m2n  by  27  m2  we  have  1  n,  which 
is  the  6  of  the  formula. 

Then  3  a2  +  3  ab  +  62  =  3(3  m)2  +  3(3  m)(i  n)  +  (1  n)2  =  27  ??i2 
+  36  ?????  +  16  n'2  is  the  complete  divisor.  This  expression  is  then 
multiplied  by  h  —  i  n,  giving  108  m2n  +  111  inn1  +  (if  »3,  which  corre- 
sponds to  6(3  a'2  +  3«6  +  62)  of  the  formula.  On  subtracting,  the 
remainder  is  zero  and  the  process  ends.  Hence,  dm  +  in  is  the 
required   root. 


78  POWERS  AND  BOOTS 

Ex.  2.     Find  the  cube  root  of 

33  x4  -  9  x'  +  xG  -  63  Xs  +  66  a?  -  36  x  +  8. 
We  first  arrange  the  terms  with  respect  to  the  exponents  of  x. 

x-  —  3  x  +  2,  cube  root 

(4iven  cube,  xG  —  9  x5  4-  ;>3  x4  —  63  s3  +  66  •'"-  —  36  x  +  8 

a3  =  re6 


3  c-  =  3  x4 
3  a2  +  3  a&  +  b2  =  3x4  -  9  x3  +  9  x2  _ 

3 a'2  =  3(«2  - 3 x)2  =  3 x4  —  18 x3      27  .■- 
3  a'2  +  3  a'6'  +  6'2  =  3  x4  -  18  x3  +  33  x2- 18  x  +  4 


—  9  x5  +  33  x4  —  63  Xs  +  m  x2  —  36  x  +  8 

—  9zs  +  27  .-■■»  —  27  x3 


6  x4--  36.  t-3  +  6i  ;./•--: 36  x      8 
6  x*  —  36  x3  +  66  .r2  —  36  x      8 


0 


The  cube  root  of  x6,  or  x2,  is  the  first  term  of  the  root.  The  first  par- 
tial divisor,  which  corresponds  to  3  a2  of  the  formula,  is  •*>(./-)'-  =  3  xi. 
Dividing—  9x5  by  3x4  we  have  —  3  x,  which  is  the  second  term  of  the 
quotient,  corresponding  to  b  of  the  formula. 

After  these  two  terms  of  the  root  have  been  found,  we  consider 
x2—Bx  as  the  a  of  the  formula  and  call  it  a'.  The  new  partial  divisor 
is  3  a'2  =  3(x2  —  3  x)2  =  3  x4  —  18  x3  +  27 x2,  and  the  new  b,  which  we 
call  &',  is  then  found  to  be  2. 

Substituting  x2  —  ■"'>  x  for  a'  and  2  for  V  in  3  a'2  +  ;1>  a'V  +  b'-,  we 
have  3  x4  —  18  x3  4-  33  a;2—  18  x  +  4,  which  is  the  complete  divisor. 
On  multiplying  this  expression  by  2  and  subtracting,  the  remainder 
is  zero.     Hence  the  root  is  x2  —  3  x  +  2. 

In  case  there  are  four  terms  in  the  root,  the  sum  of  the  first 
three,  when  found  as  above,  is  regarded  as  the  new  a,  called 
a".  The  remaining  term  is  the  new  b  and  is  called  b".  The 
process  is  then  precisely  the  same  as  in  the  preceding  step. 

EXERCISES 

Find  the  square  roots  of  the  following: 

1.  in-  +  4  inn  +  6  ml+  4  //-'  +  1 2  In  +  9  P. 

2.  4  x4  +  8  ax3  +  4  <r.c  +  1 6  tfx2  +  16  alrx  +  1 6  b\ 

3.  9  a2  -  6  ab  +  30  ac  +  6  ad  +  b-  - 10  be  -  2  bd  4-  25  c2 

+  10cd  +  d2. 


HOOTS   OF  POLYNOMIALS  79 

4.  9  a2  -  30  ab  -  3  ab2  +  25  &2+  5&3  +  -. 

4 

5.  |  era4  —  |  o6r32!  + 1  arbxrz2  +  &2<cV  —  4  ab2xz"  +  4  a2b2z4. 

6.  a2  -  6  aft  + 10  ac  -  14  ad  +  9  62  -  30  be  +  42  &d  +  25  c2 

7.  |  +  6  a-  -  17  x2  -  28  .t-3  +  49  x\  [  -  70  cd  +  49  d2. 

8.  9  a6  -  24  a3&4  -  18  aV  +  6  a\l2  + 16  &8  +  24  &V  -  8  b\l2 

9.  25  o4"7/*  -  70  a"mb5k  +  49  a'5'"&4*.  [  +  9  c10  -  6  c5d2  +  cf*. 

10.  xw  -  8  a8™*  +  16  ww  -  4  x?y3  +  16  ^  +  4  f  +  6  aV 

-  24  z4w"  -  12  y'Y  +  9  z8. 
Find  the  cube  root  of  each  of  the  following : 

11.  cc3  —  3  x2y  4-  3  xy2  —  if  +  3  a,2z  —  6  a:?/ 2  +  3  y2z  +  3  xz2 

12.  1728  a6  + 1 728  x4f  +  576  a-2/  +  64  f.  [  -  3  yz2  +  z3. 

13.  «"'  +  3  a2b  +  3  a2c  +  3  a&2  +  6  a&c  +  3  ac2  +  &3  +  3  &2c 

14.  8«3-12(r&  +  6a62-&3.  [  +  3  6c2  +  e3. 

15.  8  x6  -  36  a8  +  114  a-4  -  207  a-3  +  285  a;2  -  225  a-  + 125. 

16.  27  z'>  —  54  az5  +  63  o2z4  —  44  a¥  +  21  aV  —  6  cCz  4-  a6. 

17.  1-9  >/2  +  39  f  -  99  f  +  156  /  -  144  yw  +  64  y2. 

18.  125  a6  -  525  x*y  +  60  .r4//2  4- 1547  aty»  -  108  a-2?/4  -  1701  a//5 

-729/. 

19.  64  Z12  -  576  P  +  2160  I8  -  4320  lG  +  4860  Z4  -  2916 I2  +  729. 

20.  a6  +  6  asb  +  15  aAb2  +  20  a3&3  + 15  a2&4  +  6  a&5  +  &6. 

21.  a9  -  9  a%  +  36  a7b2  -  84  a6&3  +  126  cc'b4  -  126  a46s  +  84  as¥ 

-36a2&7  +  9a&8-&9. 

22.  a8  4-  6  a26  -  3  «2c  +  12  a&2  -  12  abc  +  3  ac2  +  8  &3  - 12  62c 

+  6  &c2  -  c3. 

23.  343  a«  -  441  a5b  +  777  a4&2  -  531  a8&8  +  444  a2&4  -  144  ab5 

+  64  b\ 

24.  als  +  12  a15  +  60  a12  +  160  a9  +  240  ofi  + 1 92  a5  +  64. 

25.  27  F~  +  189  ln  + 198  P  -  791  P  -  594  Is  +  1701 Z7  -  729  E 


80  POWERS  AND  ROOTS 

ROOTS   OF   NUMBERS  EXPRESSED   IN   ARABIC   FIGURES 

124.  The  cube  root  of  a  number  expressed  in  Arabic  figures,  as 
in  the  case  of  square  root,  pp.  225-229,  E.  C,  may  be  found  by 
the  process  used  for  polynomials.     An  example  will  illustrate. 

Ex.  1.    Find  the  cube  root  of  389,017. 

In  order  to  decide  how  many  digits  there  are  in  the  root,  we 
observe  that  103=  1000,  1003=  1,000,000.  Hence  the  root  lies  between 
10  and  100,  that  is,  it  contains  two  digits.  Since  703  =  313,000  and 
803  =  512,000,  it  follows  that  7  is  the  largest  number  possible  in  tens' 
place.     The  work  is  arranged  as  follows: 

The  given  cube.  389  017  [70  +  3,   cube  root. 

«3  =  703  =  3 13  000    1st  partial  product. 
:;„-=  3-702=  14700 
3  ab  =  3  •  70  •  3  =      630 
b2  =  32  =  9 


10  017    1st  remainder. 


3  a2  +  3  ah  +  jfl  -  15339  j  46Q17  =  b  (3  a3  +  3  ab  +  b-). 

0 

Having  decided  as  above  that  the  a  of  the  formula  is  7  tens,  we  cube 
this  and  subtract,  obtaining  46,017  as  the  remaining  part  of  the 
power. 

The  first  partial  divisor,  3  a2  =  14,700,  is  divided  into  46,017,  giving 
a  quotient  3,  which  i>  the  b  of  the  formula.  Hence  the  first  complete 
divisor,  '■'<  u-  +  Zab  +  3  &2,  is  l~>.:j:'>!t  and  the  product,  &(3  a-  +  3  ab  +  //-), 
is  46,017.  since  the  remainder  is  zero,  the  process  ends  and  7:>  is  the 
cube  root  sought. 

125.  The  cube  of  any  number  from  1  to  9  contains  one,  two. 
or  three  digits;  the  cube  of  any  number  between  10  and  W 
contains  four,  live,  or  six  digits ;  the  cube  of  any  number 
between  100  and  999  contains  seven,  eight,  or  nine  digits,  etc. 
Hence  it  is  evident  that  if  the  digits  o\'  a  number  are  separated 
into  groups  of  three  figures  each,  counting  from  units'  place 
toward  the  left,  the  number  of  groups  thus  formed  is  the  same 
as  the  number  of  digits  in  the  root. 


ROOTS    OF  ARABIC  NUMBERS 


81 


Ex.  2.    Find  the  cube  root  of  13,997,521. 

The  given  cube,  13  907  521  |200  +  40  +  1  =  241,  cube  root. 

us  =  2008  =     8  000  000 
3  a2  =  120000 


3  ab  =    24000 

b2  =      1600 

145600 

3  a'-  =  172800 
3  a'V  =       720 

b'2  = 1 

173521 


5  997  521 


5  824  000  =  b  (3  a2  +  3  aft  +  ft2) 


17:3  521 


173  521  =  V  (3  a'2  +  3  a'6'  +  b'2). 


0 


Since  the  root  contains  three  digits,  the  first  one  is  the  cube  root 
of  8,  the  largest  integral  cube  in  13. 

The  first  partial  divisor,  3  •  200-  =  120,000,  is  completed  by  adding 
3  ab  =  3  <■  200  ■  40  =  24,000,  and  b2  =  1000. 

The  second  partial  divisor,  3  a'-2,  which  stands  for  3(200  4-  40) 2 
=  172,800,  is  completed  by  adding  3  a'V  which  stands  for  3  •  240  •  1  = 
720,  and  b12  which  stands  for  1,  where  a'  represents  the  part  of  the 
root  already  found  and  b'  the  next  digit  to  be  found.  At  this  step 
the  remainder  is  zero  and  the  root  sought  is  241. 

EXERCISES 

Find  the  square  root  of  each  of  the  following: 

1.   58,081.  2.   795,564.  3.   11,641,744. 

Find  the  cube  root  of  each  of  the  following : 

4.  110,592.  7.    205,379.  10.   2,146,689. 

5.  571,787.  8.    31,855,013.  11.    19,902,511. 

6.  7,301,384.  9.   5,929,741.  12.    817,400,375. 

126.  Since  the  cube  of  a  decimal  fraction  has  three  times  as 
many  places  as  the  given  decimal,  it  is  evident  that  the  cube 
root  of  a  decimal  fraction  contains  one  decimal  place  for  every 
three  in  the  cube.  Hence  for  the  purpose  of  determining  the 
places  in  the  root,  the  decimal  part  of  a  cube  should  be  divided 
into  groups  of  three  digits  each,  counting  from  the  decimal 
point  toward  the  right. 


82 


POWERS   AXD    ROOTS 


Ex.    Approximate  the  cube  root  of  34.507  to  two  places  of 
decimals. 


a*  =  33  = 
3a2  =  3-32  =  27. 
3a&  =  3-3(.2)  =    1.8 

&2  =  (.2)-  =      .04 

L'S.SJ 

3«'2  =  3(3.2)2  =  30.72 
3a'b'  =  3(3.2)  (.06)=      .48 

6'2  =  (.05)2  =      .0025 
31.2025 
3a"2=3(3.25)2=31.6875 
3a"b"  =3(3.25)(.007)  =    .06825 
b»*  =  (.Q01)*=     .000040 


34.567  |3+  .2  +  -05  +  .007  =  3.267 
27.000 


7.507 


5.768  ft(3  a'2  +  3  aft  +  ft2) 

1.7  KOI  IU(I 


1.560125 


ft'(3a'*  +  Sa'6'  +  6'2) 


.238875000 


31.7557091     .222290503  =  6"(3a"2  +  3  a"5"+  ft"2) 
.016584407 

The  decimal  points  are  handled  exactly  as  in  arithmetic  work. 

127.  Evidently  the  above  process  can  be  carried  on  indefi- 
nitely. 3.257  is  an  approximation  to  the  cube  root  of  34.567. 
In  fact  the  cube  of  3.257  differs  from  34.567  by  less  than  the 
small  fraction  .017.  The  nearest  approximation  using  two 
decimal  places  is  3.20.  If  the  third  decimal  place  were  any 
digit  less  than  5,  then  3.25  would  be  the  nearest  approximation 
using  two  decimal  places.  Hence  three  places  must  be  found 
in  order  to  be  sure  of  the  nearest  approximation  to  two  places. 

EXERCISES 

Approximate  the  cube  root  of  each  of  the  following  to  two 


places  of  decimals. 

1.   21.4736. 

6. 

.003. 

11. 

.004178. 

2.   6.5428. 

7. 

.3017. 

12. 

200.002. 

3.   58. 

8. 

.5. 

13. 

572.271. 

4.    2. 

9. 

.05. 

14. 

31.7246. 

5.    3. 

10. 

6410.37 

15. 

54913.416. 

16.    Approximate 

the 

square 

root  in 

Exs. 

1,  2,  10,  11,  and 

15  of  the  above 

list. 

CHAPTER   VII 


QUADRATIC   EQUATIONS 
EXPOSITION    BY    MEANS    OF    GRAPHS 

128.  We  saw,  §  65,  that  a  single  equation  in  two  variables 
is  satisfied  by  indefinitely  many  pairs  of  numbers.  If  such  an 
equation  is  of  the  first  degree  in  the  two  variables,  the  graph  is 
in  every  case  a  straight  line. 

We  are  now  to  consider  graphs  of  equations  of  the  second 
degree  in  two  variables.     See  §  66. 

Ex.  1.    Graph  the  equation  y  =  x2. 

By  giving  various  values  to  x  and  computing  the  corresponding 
values  of  y,  we  find  pairs  of  numbers  as  follows  which  satisfy  this 
equation  : 

(x  =  0,  f  x  =  1,  f  x  -  -  I,  I"  x  =  2,  f  x  =  -  2,  (  x  =  3,  J  x  =  -  3,  etc> 
U  =  0.  iy  =  l.  ly  =  l.       Ly  =  4.  \y  =  4.       ly  =  9.    iy  =  9. 

These  pairs  of  numbers  correspond  to  points  which  lie  on  a  curve 
as  shown  in  Figure  3. 

By  referring  to  the  graph  the 
curve  is  seen  to  be  symmetrical 
with  respect  to  the  y-axis.  This 
can  be  seen  directly  from  the 
equation  itself  since  x  is  involved 
only  as  a  square  and  hence,  if  y  =  x2 
is  satisfied  by  x  =  a,  y  =  b,  it  must 
also  be  satisfied  by  x  =  —  a,  y  =  b. 

It  may  easily  be  verified  that 
no  three  points  of  this  curve 
lie  on  a  straight  line.  The 
curve  is  called  a  parabola. 

83 


V(-3 

9) 

+  '■> 

G 

,9) 

+  s 

CO 

+7 

+6 

% 

+8 

K-2 

l\ 

-t  l 

/' 

1) 

■c 

+2 

(-1 

1) 

■tl 

1) 

: 

_ 

! 

- 

N^ 

^?l" 

^ 

2 

-) 

3 

X 

-(/. 

is 

c 

,0) 

-1 

Fig.  3. 


84 


Q  UADRA  TIC  EQ  UA  TIONS 


■i  ■ 

ta 

..ii 

+i 

(2,5y 

+3 

v 

.» 

+2 

; 

id 

i 

_ 

2 

S> 

0 

+2 

i- 

1,0) 

X- 

IX 

s 

-2 

(u 

1 

-3> 

(ft, 

:'•) 

(-1, 

0 

-4 

| 

Fig.  4. 
we  have  the  graph  of  the  equation,  as  in  Figure  4. 


Ex.  2.    Graph  the  equation 

y  =  x-  +  2x  —  3. 

Each  of  the  following  pairs  of 
mi  ml  mt.-  satisfies  the  equation  : 

f*  =  o,     fx  =  i,  r»  =  -  i, 

J  a;  =  2,    fx  =  -2,    far  =  -3, 
[y=5.    [y  =  -  3.    ly  =  0. 

X  =  —  4, 

y  =  5. 

Plotting  these  points  and  draw- 
ing a  smooth  curve  through  them, 


EXERCISES 

In  this  manner  graph  each  of  the  following: 

1.  y  =  x2  —  1.  7.   y  =  5x  —  xr  —  4. 

2.  y  —  .r  +  4  x.  8.   ?/  =  4  x  —  ar'  +  5. 

3.  y  =  ar  +  3  a:  —  4.  9.  ?/  =  a;2  -f-  5  x  —  6. 

4.  v  =  .r  +  5  a;  +  4.  10.  y  —  —  x-  +  x. 

5.  //  =  .r  —  7x  +  6.  11.  y  =  4ic2  —  3  a;  —  1. 

6.  y  =  3x*  —  7x  +  2.  12.  y  =  —  4x- +  3x  +  l. 

129.  We  now  seek  to  find  the  points  at  which  each  of  the 
above  curves  cuts  the  aj-axis.  The  value  of  y  for  all  points  on 
the  .r-axis  is  zero.  Hence  we  put  y  =  0,  and  try  to  solve  the 
resulting  equation. 

Thus  in  Ex.  2  above,  if  y  =  0,  a:2  +  2x  -  3  =  (x  +  3)(a:  -  1)  =  0, 
which  is  satisfied  by  x  =  l  and  x=  —3.  Hence  this  curve  cuts  the  ar-axis 
in  the  two  points  sc=l,  .y  =  0  and  x=  —.'5,  //  =  0,  as  shown  in  Figure  4. 

Similarly  in  Ex.  1,  it'  y  =  0,  x-  =  0,  and  hence  x  =  0.  Hence  the 
curve  meets  the  x-axis  in  the  point  x  =  0,  y  =  0,  as  shown  in  Figure 
3.     On  this  point  sec  §  131,  Ex.  2. 


ALGEBRAIC  SOLUTION  85 

EXERCISES 

Find  the  points  in  which  each  of  the  twelve  curves  in  the 
preceding  list  cuts  the  a>axis. 

Notice  that  in  every  case  the  expression  to  the  right  of  the  equality 
sign  can  be  factored,  so  that  when  y  =  0  the  resulting  equation  in  x 
may  be  solved  as  in  §  94. 

Ex.  3.  Plot  the  curve  y  =  <Bz  +  4sc-f-2  and  find  its  intersec- 
tion points  with  the  #-axis. 

We  are  not  able  to  factor  a;2+4a;+2  by  inspection.  Hence  we  solve 
the  equation  x'2  +  4  x  +  2  =  0  by  completing  the  square  as  in  §  175, 
E.  C,  obtaining  x  =  —  2  +  a/2  and  x  =  —  2  —  a/2.  Hence  the  curve 
cuts  the  x-axis  in  points  whose  abscissas  are  given  by  these  values  of  x. 

In  making  this  graph,  we  first  plot  points  corresponding  to  integral 
values  of  x,  as  before ;  then,  in  drawing  the  smooth  curve  through 
these,  the  intersections  made  with  the  x-axis  are  approximately  the 
points  on  the  number  scale  corresponding  to  the  incommensurable 
numbers,  -  2  +  a/2  and  -  2.  -  a/2.     See  §  109. 

EXERCISES 

In  this  manner,  find  the  points  at  which  each  of  the  follow- 
ing curves  cuts  the  x--axis,  and  plot  the  curves.  For  reduction 
of  the  results  to  simplest  forms,  see  §§  159,  160,  E.  C. 

1.  y  =  x*  +  5x  +  3.  5.  y  =  2x  —  5z2  +  8. 

2.  i/  =  3x2  +  Sx-2.  6.  y  =  5  +  Sx  —  3x2. 

3.  y  =  6.x- -4 x2  +  5.  7.  //  =  3-9.r-ll.t\ 

4.  y  =  —  4-2a  +  5ar.  8.  y  =  —  2  —  2x  +  x>. 

130.  Each  of  the  foregoing  exercises  involves  the  solution  of 
an  equation  of  the  general  form  ax2  +  bx  +  c  =  0.  Obviously, 
by  solving  this  equation,  we  shall  obtain  a  formula  by  means 
of  which  every  equation  of  this  type  may  be  solved.  See 
§  179,  E.  C. 

The  two  values  of  x  are : 

Xl  2a  '  X"  2a 


86  QUADRATIC  EQUATIONS 

EXERCISES 

By  means  of  this  formula,  find  the  solutions  of  each  of  the 
following  equations : 

1.  2  a2- 3a -4  =  0.  11.  3  a; -9o*  4- 1  =  0. 

2.  3xz  +  2aj  — 1  =  0.  12.  7a?  —  3x-2  =  0. 

3.  3  sc2  —  2  x*  -  1  =  0.  13.  G  x2  +  7  a-  +  1  =  0. 

4.  4  3^4- 6  a;  4-1  =  0.  14.  4  .r  4- 5a- 8  =  0. 

5.  .r  -  7  a  + 12  =  0.  15.  4  sc2  -  5  a;  -  3  =  0. 

6.  5jc2  +  8aj4-3  =  0.  16.  S.r  +  3a-5  =  0. 

7.  5  a2  -  8  a;  +  3  =  0.  17.  7  as*  +  as  — 3  ="0. 

8.  5a24-8a-3  =  0.  18.  7k2  — a;  — 4  =  0. 

9.  5 a2 -8a;- 3  =  0.  19.  cc2-2aaj  =  3&-a2. 
10.  2a;  —  3^4-7  =  0.  20.  a2  — 6aa  =  49c2- 9«2. 

2  ,  o(o  +  6)  ,  (a +  6)  a; 

21.  x-  -\ *— ! — l  =  ax  4-  i — o         • 

3  3 

22.  —  2ar —  x  —  2  cza;  =  — * *• 

2  2 

23.  ar +  2  win  =  4  na. 

24.  a2  —  2  «a  4-  4  ab  =  b2  +  3  a2. 

25.  a*2  —  abx  4-  a26  —  ax  =  a&2  —  6a. 

26.  a24-9  — C  =  Ga. 

27.  /(.<-'2  +  urn  =  //?»2a  4-  /xa. 

28.2  (a  +  1) a2  -  (a  +  l)2a  +  2  (a  4- 1)  =  4  x. 

29.  a2  +  0  Cd  4-  3  c  =  (3  c  4-  3  d  +  1  )a. 

30.  x2  +  2  a2  4-  3  a  —  2  =  (3  a  +  1)  a 

131.    We  now  consider  the  intersections  of   other  straight 
lines  besides  the  avaxis  with  curves  like  those  plotted  above. 


DISTINCT,    COINCIDENT,   AND  IMAGINARY  ROOTS      87 

Ex.  1.  Graph  on  the  same  axes  the  straight  line,  y  =  —  2 
and  the  curve,  y  =  x2  +  2  x  —  3. 

This  line  is  parallel  to  the  .r-axis  and  two  units  below  it.  Tt  cuts 
the  curve  in  the  two  points  whose  abscissas  are  x1  =  —  1  +  V2  and 
x2  =  -  1  —  V'2,  as  found  by  substituting  —  2  for  y  in  y  =  x2  +  2x  —  3 
and  solving  the  resulting  equation  in  x  by  the  formula,  §  130. 

Ex.  2.    Graph  on  the  same  axes  y  =  —  4  and  y  =  x2  +  2  x  —  3. 
This  line  seems  not  to  cut  the  curve  but  to  touch  it  at  the  point  whose 
abscissa  is  x  =  —  1. 

Substituting  and  solving  as  before,  we  find, 

2  +  0  =  _  x 


-2+   Vi- 

1 

2 

_  2  -  V4  - 

-  4 

and  xo  =  —  — := =  —  1. 

2  2 

In  this  case  the  two  values  of  x  are  equal,  and  there  is  only  one  point 
common  to  the  line  and  the  curve.  This  is  understood  by  thinking  of 
the  line  y  —  —  2,  in  the  preceding  example,  as  moved  down  to  the  position 
y=  —  4,  whereupon  the  two  values  of  x  which  were  distinct  now  coincide. 


132.    From  the  formula,  x  = —      ,  it  is  clear  that 

2  a 

the  general  equation,  ax2  +  bx  +  c  =  0  has  two  distinct  solutions 
unless  the  expression  b2  —  4  ac  reduces  to  zero,  in  which  case  the 

two   values    of    x    coincide,    giving    xx  = ^—  = :  -   and 

^  _  _fr_Q_      j>_  2a  2a 

X-~      2  a      ~      2  a 

Ex.1.     In    2  x2  —  9  x  +  8  =  0,    determine    without    solving 
whether  the  two  values  of   x  are  distinct  or  coincident. 

In  this  case,  a  =  2,  b  =  —  9,  c  =  8. 

Hence  Ifi  —  4.  ac  =  81  —  64  =  17. 

Hence  the  values  of  x  are  distinct. 

Ex.  2.    In  4  x-2  —  12  x  +  9  =  0,  determine  whether  the  values 
of  x  are  distinct  or  coincident. 

In  this  case,  b2  —  4  ac  =  144  —  4-4-9  =  0.     Hence  the  values  of  x 
are  coincident. 


88  QUADRATIC  EQUATIONS 

EXERCISES 

In  each  of  the  following,  determine  without  solving  whether 
the  two  solutions  are  distinct  or  coincident : 

1.  .r-7.<-  +  4  =  0.  6.  6  x2  -  3  x  —  1  =  0. 

2.  4  <b*  +  28  x  +  49  =  0.  7.  4  a2  — 16  x  +  16  =  0. 

3.  9arJ  +  12a;  +  4  =  0.  8.  8  x2 -  13  =  4  x. 

4.  ;r  +  6.r  +  9  =  0.  9.  12 a2- 18  =  24 a. 

.      5.    _ic2  +  9aj  +  25  =  0.  10.    16 a?- 56 a?  =—49. 

133.  Definition.     A  line  which  cuts  a  curve  in  two  coincident 
points  is  said  to  be  tangent  to  the  curve. 

134.  Problem.     What  is  the  value  of  a  in  y  =  a,  if  this  line 
is  tangent  to  the  curve  y  =  x1  +  5  x  +  8  ? 

Substituting  «  for  ^/  and  solving  by  means  of  the  formula,  we  have 


-5±  V25-4(8-a)j 
o 

If  the  line  is  to  be  tangent  to  the  curve,  then  the  expression  under  the 
radical  sign  must  be  zero  so  that  the  two  values  of  x  may  coincide. 
That  is,  25  -  4(8  -  a)  =  0,  or  a  =  J. 

On  plotting  the  curve,  the  line  y  =  |  is  found  to  be  tangent  to  it. 

EXERCISES 

In  the  first  18  exercises  on  p.  8G  obtain  equations  of  curves 
by  letting  the  left  members  equal  y.  Then  find  the  equations 
of  straight  lines,  y  =  <t,  which  are  tangent  to  these  curves. 

135.  Problem.  Find  the  intersection  points  of  the  curve 
?/  =  .r2  +  3  x  +  5  and  the  line  y  =  21. 

Substituting  for  y  and  solving  for  x  we  have 


_  -  6  4  V36  -40       -6  +  2V-1 
Xl                   4                              4 

-3+  v 
o 

-  3  -  V 

/-  1 

^  _  _  6  -  V36  _40_-6-2V-l_ 

-  1 

DISTINCT,   COINCIDENT,   AND  IMAGINARY  ROOTS      89 

These  results  involve  the  imaginary  unit  already  noticed  in 
§  112.  Numbers  of  the  type  a  +  &V—  1  are  discussed  further 
in  §  195.  For  the  present  we  will  regard  such  results  as  merely 
indicating  that  the  conditions  stated  by  the  equations  cannot 
be  fulfilled  by  real  numbers.  This  means  that  the  curve  and 
the  line  have  no  jjoint  in  common,  as  is  evident  on  constructing 
the  graphs. 

By  proceeding  as  in  §  134  we  find  that  the  line  y  =  -^  is  tangent 
to  the  curve  y  =  x'2  +  3  x  +  5.  Clearly  all  lines  y  =  a,  in  which  a  >-jl, 
are  above  this  line  and  hence  cut  this  curve  in  two  points. 

All  such  lines  for  which  a  <  y  are  below  the  line  y  =  y  and  hence 
do  not  meet  the  curve  at  all. 

Solving  y  =  a  and  y  —  x-  -f-  3  x  +  5  for  x  by  first  substituting  a  for  y 
we  have 

-  3±  V4  a  -  11 

x  = — . 

2 

If  a  >-V"  the  number  under  the  radical  sign  is  positive,  and  there  are 
two  real  and  distinct  values  of  x.  Hence  the  line  and  the  curve  meet 
in  two  points. 

If  a<-V-)  the  number  under  the  radical  sign  is  negative.  Con- 
sequently the  values  of  x  are  imaginary  and  the  line  and  the  curve  do 
not  meet. 

Hence  we  see  that  the  conclusions  obtained  from  the  solution  of 
the  equations  agree  with  those  obtained  from  the  graphs. 

136.  From  the  two  preceding  problems  it  appears  that  it  is 
possible  to  determine  the  relative  positions  of  the  line  and  the 
curve  viitliout  completely  solving  the  equations.  Namely,  as 
soon  as  y  is  eliminated  and  the  equation  in  x  is  reduced  to 
the  form  ax2  +  bx  +  c  =  0,  we  examine  lr  —  4  ac  as  follows  : 

(1)  If  lr  —  4  ac  >  0,  i.e.  positive,  then  the  line  cuts  the  curve 
in  two  distinct  points. 

(2)  If  b2  —  4  ac  =  0,  then  the  line  is  tangent  to  the  curve. 
See  §  133. 

(3)  If  b2  —  4  ac<0,  i.e.  negative,  then  the  line  does  not  cut 
the  curve. 


90  QUADRATIC  EQUATIONS 

137.    Problem.     Find  the  points  of  intersection  of 

y  =  a?+3x  +  13  (1),    and   y+3x=7  (2). 

Eliminating  y  and  reducing  the  resulting  equation  in  x  to  the  form 
ax-  +  bx  +  c  =  0,  we  have  x2  +  6  x  +  (5  =  0. 

Solving,  a?j  =  —  3  +  VS,  x2  =  —  3  —  VVi. 

Substituting  these  values  of  z  in  (li)  and  solving  for  y,  we  have 

.r.  =  -  3  +  V3  f  ./•„  =  -  3  -  VS 

,_      and     {  ,r_ 

yi  =  16  -  3  V3  [  y2  =  lb"  +  3  V3 

which  are  the  points  in  which  the  line  meets  the  curve. 

Here  b'2  —  4  ac  =  12,  which  shows  in  advance  that  there  are  tivo 
points  of  intersection. 

EXERCISES 

In  each  of  the  following  determine  without  graphing  whether 
or  not  the  line  meets  the  curve,  and  in  case  it  does,  find  the 
intersection  points : 

(y  =  2x*-3x-4:,  6      (y  =  5x*  +  8x  +  3, 

'     [y-x  =  3.  '     [2y-5x-2  =  0. 

\    y  =  2x*  +  2x-l,  \y  =  5a?-Sx  +  3, 

\2y  =  x-l.  '     [3-x  =  3y. 

y  =  3 .<•-' -  2  x -  1,  8      [  y  =  -  5  x-  +Sx- 3, 


4. 


2x-y  =  4.  I  2-4//-.i-  =  0. 

?/  =  4.r  +  6.r  +  l,  9      \y  =  —  5x2  —  Sx  —  3, 

x  =  y  +  5.  \  5  y  —  3  a?  =  8. 


5      (y=*2-7a-  +  12,  1Q      fy  =  3a>-3a»  +  7, 

[5a;  —  2/  =  —  1.  "    \—  5  —  3x  +  2y=0. 

138.   Problem.    G-raph  the  equation  jb2  +  y2  =  25. 

Writing  the  equation   in  the  form  //  =  ±  V^o  —  x'2,  and  assigning 
values  to  x,  we  compute  the  corresponding  values  of  //  as  follow  3  : 

J  x  =  0,        J  a?  =  ±  5,     |  x  =  3,         f  x  =  -  3,      f  x  =  4,         f  z  =  -  4, 
ljf=±5.     U  =  0.         ly  =  ±4.      ly  =  ±4.      U  =  ±3.     |y  =  ±3. 


DISTINCT,    COINCIDENT,   AND   IMAGINARY  ROOTS      91 


1 

HI 

5) 

(-3,41 

JM> 

■2 

NJ(U) 

(-*,3) 

\  * 

r^\ 

ft 

>£* 

,> 

-"I 
—a 

y 

>v 

»J 

(-n.o) 

(0,0) 

xi 

(5,0) 

X- 

axis 

(ri 

,jk 

/r'' 

1) 

(-3 

-4)|V 

(M) 

(0,^5) 

FlG.   5. 


Evidently,  for  x  greater  than  5  in  absolute  value,  the  corresponding 
y's  are  imaginary,  and  for  each  x  between  —  5  and  +  5  there  are  two 
y's  equal  in  absolute  value,  but 
with  opposite  signs. 

It  seems  apparent  that  these 
points  lie  on  the  circumference 
of  a  circle  whose  radius  is  5,  as 
shown  in  Figure  5.  Indeed,  if  we 
consider  any  point  xv  yl  on  this 
circumference,  it  is  evident  that 
x\~  +  Hi2  =  -5,  since  the  sum  of 
the  squares  on  the  sides  of  a  right 
triangle  is  equal  to  the  square 
on  the  hypotenuse.  (See  figure, 
p.  207,  E.  C) 

The  equation  x-  +  y2  =  25  is, 
therefore,     the    equation    of    a 
circle  with  radius  5.      Similarly,  x'2  +  y2  =  r2  is  the  equation  of  a 
circle  with  center  at  the  point  (0,  0)  and  radius  r. 

139.  Problem.  Find  the  points  of  intersection  of  the  circle 
x2  +  y2  =  25  and  the  line  x  -f  y  =  7. 

Eliminating  y  from  these  equations,  and  reducing  the  equation  in 
x  to  the  form  ax2  +  bx  +  c  =  0,  we  have 

x2  -  7  x  +  12  =  0. 
From  which  xx  =  4,  x„  =  3. 

Substituting  these  values  of  x  in  x  +  y  =  7,  we  have  y1  =  3,  y2  =  4. 
Hence  xx  =  4,  y{  =  3  and  x9  =  3,  y2  =  4  are  the  required  points. 
Verify  this  by  graphing  the  two  equations  on  the  same  axes. 

140.  Problem.  Find  the  points  of  intersection  of  the  circle 
x2  +  y2  =  25  and  the  line  3  x  +  4  y  =  25. 

Eliminating  y  and  solving  for  x,  we  find  x  =  — ^—  =  3. 

Hence  xx  =  x2  =  3,  from  which  ?/,  =  y2  =  4. 

Since  the  two  values  of  x  coincide,  and  likewise  the  two  values  of 
y,  the  circumference  and  the  line  have  but  one  point  in  common. 
Verify  by  graphing  the  line  and  the  circle  on  the  same  axes. 


92  QUADRATIC  EQUATIONS 

141.    Problem.    Find  the  points  of  intersection  of 

x2  +  y2  =  25 
and  x  +  y  =  10. 

Substituting  for  y  and  solving  for  x  we  have 

_  20  ±  v400  -  600  _  20  ±  V-  200 
4  4 

20  ±  10  V^2      10  ±  5  v/^2" 
4  2 

The  imaginary  values  of  a:  indicate  that  there  is  no  intersection 
point.     Verify  by  plotting. 

EXERCISES 

In  each  of  the  following  determine  by  solving  whether 
the  line  and  the  circumference  meet,  and  in  case  they  do, 
find  the  points  of  intersection.  Verify  each  by  constructing 
the  graph. 

.r2  +  ,v2=16,         5      \x>  +  y2=l,  0      p2-f  ?/2=12, 

x  +  ii  =  4.  \  x  +  y  =  8.  [  x  —  y  =  6. 

x2  +  y2  =  36,         6      (.r2  +  y-  =  8,  1Q      {.r°-  +  y-  =  4, 


4. 


4  x  4-  y  =  6.  [  x  —  ?/  =  4.  {  2  x  —  3  y  =  4. 

r.r-  +  r  =  2o,  1^  +  ^  =  41,  1-^+^=40, 

\2oj  +  y=-5.  *     \x-Sy  =  7.  '     \.r  +  2,/  =  10. 

ra2  +  ?/-o  =  20,  a      {x*  +  y*=:29,  12      (.r2  +  / 


[2x  +  y  =  0.  \3x-7y=-29.    ■        \x  +  y  =  9. 


142.  Problem.  Graph  on  the  same  axes  the  circle,  x2+y2=b2, 
and  the  lines,  3 x  +  4y  =  20,  3  a  +  4  //  =  25,  and  3 sc  +  4ty  =  30. 

The  first  lino  cuts  the  circumference  in  two  distinct  points,  the 
second  seems  fco  be  tangent  to  it.  and  the  third  does  not  meet  it.  Ob- 
ser\e  that  the  three  lines  are  parallel.     See  Figure  0. 


DISTINCT,    COINCIDENT,  AND  IMAGINARY  ROOTS      93 


In  order  to  discuss  the 
relative  positions  of  such 
straight  lines  and  the  circum- 
ference of  a  circle,  we  solve 
the  following  equations  simul- 
taneously : 

x2  +    y2  — 1~  (1) 

3x  +  ±y  =  c  (2) 

Eliminating  y  by  substitution, 
and  solving  for  x,  we  find 


s  "In 

V 

IflJUS. 

v   rv^ 

<&$* 

» 

1  X 

•2 

L^^n!^ 

8 

It 

(- 

-..111 

* 

(0,0) 

(5,0) 

X- 

axis 

CO,- 

5) 

3  c  ±  4  V-25  r2  -  c2 


(3) 


Fig.  6. 


The  two  values  of  x  from  (3)  are  the  abscissas  of  the  points  of 
intersection  of  the  circumference  (1)  and  the  line  (2). 

These  values  of  x  are  real  and  distinct  if  25  r2  —  c2  is  positive,  real 
and  coincident  if  25  r2  —  e2  =  0,  and  imaginary  if  25  r2  —  c'2  is  negative. 

Now  25  r2  —  c2  is  positive  if  r  =  5,  c  =  20;  zero  if  r  =  5,  c  =  25 ;  and 
negative  if  r  =  5,  c  =  30. 

Hence  these  results  obtained  from  the  solution  of  the  equations 
agree  with  the  facts  observed  in  the  graphs  above. 

143.  Definition.  Letters  such  as  c  and  r  in  the  above  solution 
to  which  any  arbitrary  constant  values  may  be  assigned  are 
called  parameters,  while  x  and  y  are  the  unknowns  of  the 
equations. 

EXERCISES 

Solve  each  of  the  following  pairs  of  equations. 

Give  such  values  to  the  parameters  involved  that  the  line  (a)  may 
cut  the  curve  in  two  distinct  points,  (b)  may  be  tangent  to  the  curve, 
(c)  shall  fail  to  meet  the  curve. 


1. 


|  X2  +tf  =  4:, 

\  ax  +  3  y  =  16. 

x2  +  y2=16, 
2  x  +  by  =  12. 


(x2  +  y2  =  25, 
\2x  +  Zy  =  c. 

y-  =  8  X, 

3x  +  4y  =  c. 


94 


Q  UADRA  TIC  EQ  UA  TIOXS 


(5tf  =  2px, 

J  x  +  y  =  c. 

f  y  =  xr  -f  maj  +  /», 
j  aj  +  ?/  =  4. 

f  ?/  =  ma;2  —  nx  —  4, 
{  x  —  3  y  =  8. 

(iy  =  2ar!-3a:  +  l, 
\  2  a;  -  by  -  1  =  0. 


10. 


11. 


12. 


f  y  =  3  a;-  +  ??ix  —  m, 

U','  +  2y+l=0. 

//  =  ma?  +  2  »a:, 
2  //  -  6a;  -5  =  0. 

(  ?/  =  .r  +  //.>;  +  1, 
{  ax  + -J  i,  =  10. 

|  iB2  +  r  =  r2, 
(  ax  -f-  by  =  c. 


144.    Problem.    Graph  the  equation   —  +  -1 


=  1. 


25      16 

Writing  the  equation  in  the  form  y  =  ±  f  V25  —  x-,  and  assigning 
values  to  a:,  we  compute  the  corresponding  values  of  y  as  follows : 

x  —  0,  I"  x  =  ±  5,  fx  =  l,  f  .<•  =  —  1, 


.'/  =  ±  4, 


2 


y  =  0,  {  //  =  ±  3.9,  =  ±  3.9. 

x  =  3,  f  x  =  4, 

2/  =  ±3.2,  U  =  ±2.4. 

Evidently  if  x  is  greater  than  5 
in  absolute  value,  the  correspond- 
ing values  of  y  are  imaginary. 

Plotting  these  points,  they  are 
found  to  lie  on  the  curve  shown  in 
Figure  7.  This  curve  is  called 
an  ellipse. 

EXERCISES 

Solve  the  following  pairs 
of  equations. 

In  this  way  determine  whether 
the  straight  line  and  the  curve 
intersect,   and   in   case   they  do, 

determine  the  coordinates  of  the  intersection  points.     Verify  each  by 

constructing  the  graphs. 

'£  +  £  =  !,  o         te  +  jd-1, 


(0,1) 

'": 

^.J.*) 

■4.2 

*)f 

H 

,(! 

-  1) 

1 

(-5I0) 

JM) 

(5,0^ 

i x 

axis 

( 

1.--' 

d» 

',| 

■2.4) 

( 

& 

3,7 

(0,-4) 

Fig.  7. 


L6      9 

3a;  +  4y  =  12. 


2. 


49     16 
2  a?  —  7  y  = ! 


DISTINCT,  COINCIDENT,  AND  IMAGINARY  BOOTS     95 


3. 


5. 


x2  +  4  y-  =  25, 
2  x  -y  =  L 

8a»+2jf=ll, 

x  —  3y  =  7. 


6. 


7.    J 


1, 


\  st+yl 

25      9 


2*-y=14 


y=2a£— 3as+4, 
y— 4a— 8=0.        9. 

1-^  +  ^  =  16, 

I x + y = 7- 

64     12  10. 

4y-2a-=4. 


36     45 

-5  x+6y  =  10. 


[-  +  ^-  =  1, 
49      25 

la  +  y-12. 


When  arbitrary  constants  are  introduced,  in  the  equations  of  a 
straight  line  and  an  ellipse,  we  may  determine  values  for 
these  constants  so  as  to  make  the  line  cut  the  ellipse,  touch  it, 
or  not  cut  it,  as  in  the  case  of  the  circle,  §  142. 


EXERCISES 

Solve  each  of  the  following  pairs  simultaneously. 

Give  such  values  to  the  constants  that  the  line  shall  (a)  cut  the 
curve  in  two  distinct  points,  (b)  be  a  tangent  to  the  curve,  (c)  have 
no  point  in  common  with  the  curve. 

In  case  (b)  is  found  very  difficult,  this  may  be  omitted. 


1. 

k2+^=i, 

a2     16 
{8x  +  5y=4Q. 

5. 

16  +  25        ' 
[  ax  +  Ay  =20. 

9. 

'  a?8  +  ?/2  =  r2, 
ax  —  3  y  =  4. 

2. 

•j  25      b- 
[4x4-15?/ =60. 

6. 

f  -r~  _l  y2    1 

i36  +  16  =  1' 

10. 

'5x2+Sy2=W, 
hx  —  ky  ==  8. 

[  ax  +  6  y  —  60  = 

0. 

3. 

1  *-  +  -2L  =  l, 

25     16 

[  4  x  —  5  y  =  c. 

7. 

\  v2   i    if      -. 
.36  +  25"1' 
[5x  +  by  =  60. 

11. 

x2  +  7?/2  =  144, 
aa5  +  6y  =  12. 

4. 

[  or      i^_1 
16  +  25        ' 

8. 

j  a      lr 

12.  ■ 

ar  +  4  y*  =  r-, 

[5x-by  =  20. 

[bx-2y  =  5. 

_  ax  +  by  =  c. 

96 


QUADRATIC  EQUATIONS 


SPECIAL   METHODS   OF    SOLUTION 

145.  We  have  thus  far  solved  simultaneously  one  equation 
of  the  second  degree  with  one  of  the  first  degree.  After  sub- 
stitution each  has  reduced  to  the  solution  of  an  ordinary  quad- 
ratic, namely,  of  the  form,  ax2  +  bx  +  c  =  0. 

While  this  is  an  effective  general  method,  yet  some  im- 
portant special  forms  of  solution  are  shown  in  the  following 
examples : 

x2  +  f  =  a,  (1) 

x  -  y  =  b.  (2) 


Ex.  1.    Solve 


Square  both  members  of  (2)  and  subtract  from  (1). 

2  xy  =  a  —  b'2. 
Add  (1 )  and  (3) .     x2  +  2 xy  +  >/2  =  2  a-  ft2. 
Hence  x  +  y  =  ±  V'2  a  —  b-. 

From  (2)  and  (5),  adding  and  subtracting 


V2  a  -  b-  +  b 


and  • 


V2  a  -  V2  -  b 


F  = 


-V2q-6a-6 

2 


Ex.  2.    Solve  \  , 

[x-y  =  b. 

From  (1)  ( ./■  -  ?/)(.r  -f  //)  =  a. 

Substituting  b  for  x  —  y  in  ('•]),    x  +  y  =  -  • 

ft 

Then  (2)  and  (1)  may  be  solved  as  above. 


Ex.  3.    Solve 


x  +  y  =  a, 


xy  =  b. 

Multiply  (2)  by  1.  subtract  from  the  square  of  (1).  and  get 
x2  —  2  xy  +  >/-  =  a'2  -  1  6, 


w  hence,  a;  —  y  =  ±  \  a2  —  46. 

Then  (1)  and  (1)  may  be  solved  as  in  Ex.  1. 


(3) 
(1) 
(5) 


(1) 
(2) 
(3) 
(4) 


(1) 

(2) 

(;5) 
(1) 


SPECIAL   METHODS   OF  SOLUTION 


97 


The  equations  i  , 

1  I  xy  =  b, 

may  be  solved  in  a  similar  manner. 

146.  We  are  now  to  study  the  solution  of  a  pair  of  equations 
each  of  the  second  degree.     See  §  66. 

Consider  x2  +  y  =  a,  (1) 

x  +  tf  =  b.  (2) 

Solving  (1)  for  y  and  substituting  in  (2)  we  have, 
x  +  a2  —  2  ax2  +  x*  =  b, 
which  is  of  the  fourth  degree  and  cannot  be   solved  by  any 
methods  thus  far  studied.     There  are,  however,  special  cases  in 
which  two  equations  each  of  the  second  degree  can  be  solved 
by  a  proper  combination  of  methods  already  known. 

147.  Case  I.      When  only  the  squares  of  the  unknowns  enter  the 
equations. 

Example.     Solve  f ^  +  b/\=  °u 
[a2x-  +  b2y  =  c2. 

These  equations  are  linear  if  x2  and  y2  are  regarded  as  the 
unknowns. 

Solving  for  a;2  and  y2  as  in  §  73,  we  obtain, 

''A      ,.i  __  a\c-y  —  02Ci 


Hence,  taking  square  roots, 


y 


ai^2  —  °2^1 


2  -  CA^ 
axb.2  —  a2bj 


-  JaiC2  ~  a?Pl} 

*  ajb2  —  a2b^ 

-  _  JCA  -~<iA 

^al6i-a8&1, 


..  -JcA  -CA 

a  A  -  «26i ' 


\ 


'  ajb2  —  aj)x 


=  Al 


^A     T2 


r/.c,  —  a„c. 


axb.2  -  a2b^ 


c2bt 
axb2  —  a.pi 


y* 


VS 


axb.,  —  a2/)j 

In  this  case  there  are  four  pairs  of  numbers  which  satisfy  the  two  equa- 
tions.   This  is  in  general  true  of  two  equations  each  of  the  second  degree. 


98 


QUADRATIC  EQUATIONS 


y\s" 

1-, 

.1, 

■W/ 

(k 

.  a\ 

/ 

"N 

// 

Iff 

\\ 

\ 

/ 

|      /  .r 

-axis 

(0,0) 

\ 

\    V— 

■2 

\\ 

A  V  V 

e 

i/l/l  / 

/ 

\ 

s> 

/  /  y 

/ 

Lfl 

:.'  ^Ov. 

^s{y 

v; 

> 

^x > 

^>J__ 

j i 

^j! 

Example.  Solve  simul- 
taneously, obtaining  re- 
sults to  one  decimal 
place : 


(1) 
:  25.        (2) 


36     16 


Fig.  8. 


Clear  (1)  of  fractions  and 
proceed  as  above.  Verify 
the  solution  by  reference 
to  the  graph  given  in 
Figure  8. 


EXERCISES 


Solve  simultaneously  each  of  the  following  pairs  of  equa- 
tions and  interpret  all  the  solutions  in  each  case  from  the 
graph  in  Figure  8  : 


36 

16 

=  1, 

.<•-' 

+  if  = 

36. 

r     9 
.1- 

36 

16 

=  1, 

.r 

+  f  = 

16. 

3. 


X~ 

36 

0 

16 

=  1, 

X2 

+  if  = 

49 

'  X2 

36 

9 

16 

=  1, 

ar 

+  f  = 

9. 

2. 


148.    Problem.     Graph    the   equation — 

25      16 


1. 


Writing  the  equation  in  the  form  y  —  ±^Vx-  —  25,  and  assigning 
values  to  z,  we  compute  the  corresponding  values  of  y  exactly  or 
approximately  as  follows: 

\x=±5,  \x  =  6\,     (x=-6\,  ix  =  7,  fx=-7,      fx=8,      fa;=-8, 

b  =  0,      U=±3,ly=±3,    \y=±S.9,   \y=±3.9,  {  y=  ±5,  \y=±5. 


SPECIAL   METHODS    OF  SOLUTION 


99 


Evidently  when  x  is  less  than  5  in  absolute  value,  y  is  imaginary, 
and  as  x  increases  beyond  8  in  absolute  value,  ij  continually  increases. 

Plotting  these  points,  they  are  found  to  lie  on  the  curve  as  shown  in 
Figure  9.     This  curve  is  called  a  hyperbola. 


( 

-S,5)j\ 

'h 

yjcs.o 

(-7 

,3.9) 

\ 

1 

(7,3 

9) 

(-6) 

i,*r 

r^,\::,) 

(  5,( 

) 

(0,0 

(5,0 

.r-( 

xis 



. 

( 

-6*. 

3V 

V(CK,-3) 

(-7,- 

3.9) 

,(V 

S.9) 

(- 

3.-5>J/ 

(8,-5) 

^- 

Fig.  9. 

EXERCISES 

Solve  each,  of  the  following  pairs  of  equations. 

Construct  a  graph  similar  to  the  one  in  Figure  8  which  shall  contain 

the  hyperbola  ^  —  ^—  =  1  and  the  circles  given  in  Exs.  1,  2,  and  3. 

Construct  another  graph  containing  the  same  hyperbola  and  the 
ellipses  given  in  Exs.  4,  5,  and  6.  From  these  graphs  interpret  the 
solutions  of  each  pair  of  equations. 


25      16 

x2  +  y-  =  16. 


25     16 

a*  +>/==  25. 


x      T  —-i 


L6 


\a?  +  y*  =  36. 


100 


QUADRA  TIC  EQUA  TIONS 


*.      J!-1 
25     16"   ' 

-^  +  ^  =  1. 
36     16 


a;' 

25 


L6 


r  =  l, 


25 


^+-?/"=l. 
25     16 


95      i«        ' 

7  +  ' 


16      9 


16 


7.    Graph  the  equation  xy  =  9. 

( >  raph  a;;/  =  8  on  the  same  axes  with  each  of  the  following : 


8.   x2  +  y-  =  16.       9.   x2+y2  =  25. 


11. 

25      16        ' 

12. 

25      10.24 

14. 

x2      y2  _  1 
25     16 

15. 

x2      y2      1 
25      9 

10.    x2  +  /  =  4. 

13.    ^  +  ^  =  1. 

16      4 


16. 


16 


£  =  1. 

4 


17.  From  those  graphs  in  Exs.  8  to  16,  in  which  the  curves- 
meet,  determine  as  accurately  as  possible  by  measurement  the 
coordinates  of  the  points  of  intersection  or  tangency. 

18.  Solve  simultaneously  the  pairs  of  equations  given  in 
Exs.  8  to  10,  after  studying  the  method  explained  in  §  150, 
Ex.  1.  Compare  the  results  with  those  obtained  from  the 
graphs. 

19.  Solve  Exs.  11  to  16  by  the  method  explained  in  §  149, 
and  compare  the  results  with  those  obtained  from  the  graphs. 

149.  Case  II.  When  all  firms  containing  the  unknowns  are  of 
the  second  degree  in  the  unknovms. 


Example.    Solve 


2x2-3xy  +  ±y2  =  3, 


3  x2  —  4  xy  +  3  y2  =  2. 
Put  y  =  vx  in  (1)  and  (2),  obtaining 


"U2(3-4i<  +  3r2)=2. 
Hence  from  (3)  and  (4), 
3 


2  -  3  v  +  4  v2 


,  and  also  x'2 


3  —  4  v  +  3  v'2 


(1) 

(2) 


(3) 
(4) 


(5) 


SPECIAL   METHODS   OF  SOLUTION 

101 

From  (5) 

3                          2 

(6) 

2  -3u  +  4i>23-4v  +  3  v2' 

»2  —  6  w  +  5  =  0. 

(7) 

Hence 

v  =  1,   and   f  =  5. 

(8) 

From  y  = 

vx, 

y  =  x,   and   y  =  5  x. 

(9) 

If  #  =  x,  then  from  (1)  and  (2), 


y  =  i, 

If  ?/  =  5  a:,  then  from  (1)  and  (2), 
1 


1  ,     f  x  =  —  1. 

'    and    ■{  ' 


12/ 


y 


V29 


V29 


and    < 


X  = 

i 

V29 

y  = 

5 
V29 

Verify  each  of  these  four  solutions  by  substituting  in  equations 
(1)  and  (2). 

150.  There  are  many  other  special  forms  of  simultaneous 
equations  which  can  be  solved  by  proper  combination  of  the 
methods  thus  far  used.  Also,  many  pairs  of  equations  of  a 
degree  higher  than  the  second  in  the  two  unknowns  may  be 
solved  by  means  of  quadratic  equations. 

The  suggestions  given  in  the  following  examples  illustrate 
the  devices  in  most  common  use. 

The  solution  should  in  each  case  be  completed  by  the  student. 


3^  +  ^  =  58, 


Ex.  1.   Solve 

Adding  twice  (2)  to  (1)  and  taking  square  roots,  we  have 


./■  +  y  =  10,   and   x  +  y  =  —  10. 


(1) 
(2) 


(3) 


Each  of  the  equations  (3)  may  now  be  solved  simultaneously  with 
(2),  as  in  Ex.  3,  p.  96. 


102 


QUADRATIC  EQUATIONS 


Ex.  2.   Solve 


M  =  5, 

x      y 

4  +  ^  =  13. 


Let    -  =  a    and    -  =  b.     Then  these  equations  reduce  to 
x  y 

f  a  +  b  -  5, 
1  a2  +  b2  =  13. 

(3)  and  (4)  may  then  be  solved  as  in  Ex.  1,  p.  96. 

Ex.3.    Solve  (x*  +  f  +  x  +  y  =  8, 

la  7/  =  2. 

Add  twice  (2)  to  (1),  obtaining 

x2  +  2  xy  +  y'2  +  x  +  y  =  12. 

Let  x  +  y  =  a.     Then  (3)  reduces  to 

a2  +  a  =  12, 
or,  o  =  3,  a  =  —  4. 

Hence  x  +  ?/  =  3,    and  x  +  y  —  —  4. 

Now  solve  each  equation  in  (5)  simultaneously  with  (2). 


Ex.  4.    Solve 


xY  +  xY  =  272, 

X*  +  y*=10. 


(1) 

(2) 


(3) 
(4) 


(1) 
(2) 


(3) 


(4) 
(«) 


(1) 

(2) 


In  (1)  substitute  a  for  z2#2.     Then 

a2  +  a  =  272,   whence   o  =  16,   and    —  17. 

Hence  xy  =  ±  VT6  =  ±  4,   and    ±  V-  17. 

Each  of  these  equations  may  now  be  solved  simultaneously  with 
(2),  as  in  Ex.  1,  p.  101. 


Ex.  5.    Solve 


•  jgS  _  ?/3  _  117j 

.  a;  -  y  =  3. 


(1) 

(2) 


SPECIAL    METHODS   OF  SOLUTION  103 

By  factoring,  (1)  becomes 

(x-yX*  +  xy  +  y*)=117.  (3) 

Substituting  3  for  x  —  y,  we  have 

x2  +  xy  +  y2  =  39.  (4) 
(2)  and  (4)  may  now  be  solved  by  substitution  as  in  §§  140-144. 

Ex.6.    Solve  K  +  /  =  513,  (1) 


x  +  y=9.  (2) 

Factor  (1)  and  substitute  9  for  x  +  y.     Then  proceed  as  in  Ex.  5. 

Ex.7.    Solve  (.^  +  ^  =  126,  (1) 

\x  +  !,  =  9.  (2) 

Factoring  (1)  and  substituting  9  for  x  +  y,  we  have 

ay  =  14.  (3) 

(2)  and  (3)  may  then  be  solved  as  in  Ex.  3,  p.  96. 

Ex.8.    Solve  (^  +  2^  =  54^,  (1) 

U  +  y  =  6.  (2) 

Factor  (1)  and  substitute  6  for  x  +  y,  obtaining 

x2  —  xy  +  y2  =  9  xy.  (3) 

(2)  and  (3)  may  now  be  solved  by  substitution,  as  in  §§  140-144. 

Ex.9.    Solve  (**-»*  =  63,  (1) 

I  a?  +  xy  +  y2  =  21.  (2) 

Factor  (1)  and  substitute  21  for  x2  +  xy  +  y2,  then  proceed  as  in 
Ex.  8. 

Ex.10.    Solve  ( 3s  +  9s  =  243.  (1) 

UV  +  .wr  =  162.  (2) 

Multiply   (2)  by  3  and    add  to    (1),  obtaining   a    perfect   cube. 
Taking  cube  roots,  we  have 

x  +  y  =  9.  (3) 

(1)  and  (3)  are  now  solved  as  in  the  preceding  example. 


104  QUADRATIC  EQUATIONS 

Ex.  11.    Solve  K  +  .7<  =  641,  (1) 

[x  +  y  =  7.  (2) 
Raise  (2)  to  the  fourth  power  and  subtract  (1),  obtaining 

4  x3y  +  6xV  +  4  xy3  =  1760.  (3) 

Factoring,       2  xy  (2  x2  +  3  xy  +  2  y-)  =  1760.  (4) 

Squaring  (2)  we  have 

o  x2  +  4  x>/  +  2  y2  =  98,  (5) 

or  2  x2  +  3  ./vy  +  2  y-  =  98  -  x#.  (6) 

Substituting  (6)  in  (4),  we  have 

2  xy  (98  -  xy)  =  1760,  (7) 

or  x°-y2  -  98  x,y  +  880  =  0.  (8) 

In  (8)  put  xy  =  a,  obtaining 

tt2  _  gg  a  +  880  =  0.  (9) 

The  solution  of  (9)  gives  two  values  for  xy,  each  of  which  may  now 
be  combined  with  (2)  as  in  Ex.  3.  p.  UG. 

EXERCISES 

Solve  each  of  the  following  pairs  of  equations  : 


r  +  rs  +  sr 
r  —  s  =  3. 


63, 


5. 


|  x2  +  y-  =  a, 
{  xy  =  b. 


3  a? +  2  y2  =  35, 
2x*-3y2  =  6. 

(3  x*  +  2  xy  =  16, 

1  4  .r  -  3  xy  =  10. 


7. 


4. 
13. 
14. 


8. 


C  a2  -f-  db  +  &2  =  ~, 
(  a2  _  a6  +  62  =  19. 

3  a -2?/ =  6, 

;i.r_  2. -7/  +  4/  =  12 

a  +  6  +  aft  =  11, 
(a  +  6)2  +  a262  =  Gl. 


9. 


10. 


11. 


12. 


,-<  +  y5  =  91, 
x  + .?/  =  7. 
x-  +  ?/2  =  a, 
x2  -y-  =  b. 
.r-—  '.)  xy  =  0, 
:,,■- +  .sy2  =9. 

i  +  i  =  W, 

1  +  1=1. 


15. 


s+A  +  i  =  49, 

CT       ah       b~ 

1+1=8. 

a      6 


SPECIAL   METHODS   OF  SOLUTION  105 

^4a2-2o&  =  &2-16,  27>     f  (x-  4)2  +  G/  +  4)2  =  100, 

5a2  =  7a6  —  36.  }a+y  =  14. 


17. 


19. 


21. 


22. 


3a2-9?/2  =  12,  fay +  2/ -ha  =  17, 

2x-3y=  14.  {  a?y  +  f  +  a2  =  129. 


18.      l-'-  +  ^/  +  r  =  «,  29      f5  +  a2  =  o  («-&), 

x2  +  y  =  6.  |  a  +  52  _  o  (a  _  6^ 


rf  +  jf  +  a  +  y-18,  r(13.,)2  +  2//2  =  177, 

^  =  6-  1(2^-13^  =  3. 


j  x2  +  y2  +  a;  —  ?/  _  36,  r/Q\2      /or;\2 


31. 


9      25 


fx-2-5a7/  +  ?/2=  -2, 

{■^  +  7.^  +  r  =  22.  i*"    y 

«2  +  6a^  +  &2  =  124,  32.     ja'2  +  .?/2  =  20> 
a  +  6  =  8.  [5.r-3y/2  =  28. 

a2_3r(6  +  2&2  =  0,  33.     f^=-6-3ajy, 


23-     J2«2  +  a&-&2  =  9.  '     l-2^  =  /-24. 

24      (x-  +  f  +  2x  +  2y  =  27,     34      (  a  +  y  +  Va~+2/  =  12, 
xy  =  -12.  "     1^  +  ^  =  189. 

a2  +  r  -  5  x  —  5  y  =  -  4,   35      f  -«4  +  x2y2  +  ;'/4  =  133, 
xy  =  5.  1  a2  -  xy  +  y2  =  7. 

26.      ((7+*)(6  +  »)  =  80»  36.     (*  +  *#  +  // =  29, 

\x  +  y  =  5.  {  x2  +  ay  +  y2  =  61. 

37      (2x*-5xy  +  3x-2y  =  22,         3g      (x  +  y  =  U, 

\5xy  +  7x-8y-2  x2  =  8.  {  x2  +  //2  =  3026. 

39      f7//2-r,.r'  +  20.r  +  13//-29, 
{  5  (x  -  2)2  -  7  y2  -  17  //  =  - 17. 


106  QUADRATIC  EQUATIONS 

40      ((3x  +  4y)(7x-2y)+3x  +  4y  =  U, 
\(3x  +  4:y)(7x-2y)-7'x  +  2y  =  30. 

41.     p  +  ."  =  4> 

1  x>  +  x*y  +  afy2  +  aft/3  +  a*/4  4-  /'  =  364. 

42.    faj5-2/!=31J  44.    (vl+y*—xy=80, 

1  •'•  -  ?/  = 1  •  I « — y  -  xy  =  -  8. 

43      fa* +  2/*  =  82,  45      r8a  +  8&-a&-a2  =  18, 

{  x2  +  t/2  +  2  x2y2  =  28.  [5a+5&  —  ft2-  a&  =  24. 


46. 


47. 


48. 


(a3  +  xh,  +  xtf  +  f)  (x  +  y)  =  325, 
(or*  -  x*y  +  xyz  —  f)  (x  —  y)  =  13. 

2(^  +  4)2-50/-7)2  =  7o, 
7  (x  +  4)2  +  15  (?/  -  7)2  =  1075. 

(as  +  tf  =  (a  +  b)(x-y), 

\  x-  —  xy  +  //"  =  a  —  b. 


HIGHER   EQUATIONS  INVOLVING   QUADRATICS 

151.  An  equation  of  a  degree  above  the  second  may  often  be 
reduced  to  the  solution  of  a  quadratic  after  applying  the  factor 
theorem.     See  §  92. 

Example.     Solve        2  x^+x2  -  10  .r  +  7  =  0.  (1) 

By  the  factor  theorem,  x  —  1  is  found  to  be  a  factor. 

giving  (x  -  1)  (2  x2  +  8  x  -  7)  =  0.  (2) 

Hence  by  §  22,  x  -  1  =  0   and   2  x1  +  3  a;  -  7  =  0.  05) 

From  x  -  1  =  0,  x  =  1.  (4) 

From  2  ./■-  f  3  x  -  7  =  0,        x  =  ~  3  ±  vfi5  .  (5) 

4 

Hence  (4)  and  (5)  give  the  three  roots  of  (1). 


HIGHER   EQUATIONS  INVOLVING   QUADRATICS      107 
EXERCISES 

Solve  each  of  the  following  equations : 

1.  7  a-3  -  liar +4  a- =  0.  5.    28  a;3-  10  x2-U  x  =  6. 

2.  3  a-4+ a-3  4- 2  ar  4- 24  a;  =  0.  6.    a-4  -3.x-3  +  3 x2 -  x  =  0. 

3.  3a-3-16ar+23a,--6=0.  7.    4  a,-8  +  12a2-  3  x  -  9=0. 

4.  5  a-3  4- 2  ar  4- 4  a- =-7.  8.    a-4  -  5  a;3  4- 2  ar  4- 20  a- =24. 

9.    6 a,-3  4- 29 ar- 19 a;  =  16. 
10.    15  .e4  4- 49  a;3 -92  a,-2  4- 28  a;  =  0. 

EQUATIONS  IN   THE  FORM  OF  QUADRATICS 

152.  If  an  equation  of  higher  degree  contains  a  certain 
expression  and  also  the  square  of  this  expression,  and  involves 
the  unknown  in  no  other  way,  then  the  equation  is  a  quadratic 
in  the  given  expression. 

Ex.  1.    Solve  xA  +  7r  =  44.  (1) 

This  may  be  written,  (x2)2  4-  7(.r2)  =  44,  (2) 
which  is  a  quadratic  in  x1.     Solving,  we  find 

x2  =  4  and  x2  =  -  11.  (3) 


Hence,  x  =  ±  2  and  x   =  ±  V-ll.  (4) 


Ex.2.    Solve  a-  +  2  4-3VaT+2  =  18.  (1) 


Since  x  +  2  is  the  square  of  Vx  +  2,  this  is  a  quadratic  in  Vx  +2. 


Solving  we  find         Vx  +  2  =  3  and  Vx  +  2  =  —  6.  (2) 

Hence  x  4-2  =  9  and  £+2=36,  (3) 

Whence  x  =  7  and  x  =  34.  (1) 

Ex.3.    Solve    (2x2-l)2-r>(2a-2-l)-14  =  0. 

First  solve  as  a  quadratic  in  2x-  —  1  and  then  solve  the  two  result- 
ing quadratics  in  x. 


108  QUADRATIC  EQUATIONS 


Ex.  4.    Solve   or  -  7  x  +  40  -  2  Vo2  -7x  +  09  =  -  20.        (1) 
Add  29  to  each  member,  obtaining 


x2  -  7  x  +  09  -  2Vx2-7x  +  69=  3.  (2) 


Solve  (2)  as  a  quadratic  in  Vx2  —  7x  +  69,  obtaining 


Vx'2  -  7a;  +  69  =  3  and  Vx2  -  7  x  +  69  =  -  1,  (3) 

whence  x2  —  7  x  +  69  =  9  or  1 .  (4) 

The  solution  of  the  two  quadratics  in  (i)  will  give  the  four  values 
of  x  satisfying  (1). 

EXERCISES 

Solve  the  following  equations  : 


1.    x6  +  2.^  =  80.  2.    o.r-4-2V5.r-4  =  03. 

3.    (2  —  a;  +  x2)2  +  x2—  x  =  18. 


4.  a2-3a+-4-3Va2-3a  +  4=  -2. 

5.  3  a6- 7  a8- 1998  =  0. 


6.    .r  - 8. r  + 16  +  GVar'- 8  a  +  10  =  40. 

7-  (a+I)!+4(a+«)=2L 

8.   a8 -97  a4 +  1296  =  0. 


9.    a2  -  3  «  +  4  +  Va2  —  3  a  +  15  =  19. 
10.    (5  x  -  7  +  3  .r2)2  +  3  a2  +  5  .v  -  247  =  0. 


11.    V7  a  -  6  -4  V7  a-  6  +  4  =  0. 

RELATIONS    BETWEEN   THE   ROOTS    AND   THE    COEFFICIENTS   OF   A 

QUADRATIC 

153.    If  in  the  general  quadratic,  ax2  +  bx  +  c  =  0,  we  divide 

b  c 

both  members  by  a  and  put  -  =p,    -  =  q,  we  have  xi-\-px-\-q=0. 

a  a 


ci-                 —  V  +  Vp2  —  4  a         ,    «      —  »  —  V/r  —  4  q 
Solving,  x,  =  — i — ! 1 i  ,  and  xr  =  — ±- — ±- 2  . 


RELATIONS    OF  COEFFICIENTS  AND  ROOTS        109 

2  n 
Adding  «i  and  x2,  x1  +  x2  =  — ~  =  —p.  (1) 

Multiplying  x1  and  x2,        x1x.2  =  ^——s^—~ — ^  =  q.  (2) 

Hence  in  a  quadratic  of  the  form  x2 -\- p>x -\- q  =  0,  the  sum  of 
the  roots  is  —  p,  and  the  product  of  the  roots  is  q . 

rP1  •         o       a  b2      4  c      b2  —  4  ac 

lne  expression  p  —  4  5  = = . 

a2      a  a2 

Hence  p2  —  4  g   is   positive,  negative,  or   zero,   according   as 

b2  —  4  ac  is  positive,  negative,  or  zero. 

Hence,  as  found  on  pp.  87,  89,  the  roots  of 

ax2  +  bx  +  c  =  0,  or  x2  +  p.x*  +  q  =  0  are : 

reaZ  and  distinct,  ifb2  —  4  ac  >  0,      or  p2  — •  4  g  >  0,  (3) 

reaZ  and  equal,  if    b2  —  4  ac  —  0,      or  p2  —  4  g  =  0,  (4) 

imaginary,  if           b2  —  4  ac  <  0,      or  j>2  —  4  g  <  0.  (5) 

By  means  of  (1)  to  (0),  we  may  determine  the  character  of 
the  roots  of  a  quadratic  without  solving  it. 

Ex.  1.    Determine  the  character  of  the  roots  of 
8  a:2  -  3  x  -  9  =  0. 

Since  b2  —  4  ac  =  9  —  4  •  8(  —  9)  =  297>0,  the  roots  are  real  and  dis- 
tinct.    Since  b'2  —  4  ac  is  not  a  perfect  square,  the  roots  are  irrational. 

Since  q  =  —  f  =  x,z2,  the  roots  have  opposite  signs. 

Since  p  =  —  |  or  —  p  =  f  =  xx  +  x2,  the  positive  i-oot  is  greater  in 
absolute  value. 

Ex.  2.  Examine  3  a-2  +  5  »  +  2  =  0. 

Since  b'2  —  4  ac  =  25  —  4  •  3  •  2  =  1  >  0,  the  roots  are  ?-eoZ  and  distinct. 

Since  62  —  4  «c  is  a  perfect  square,  the  roots  are  rational. 

Since  q  —  §  =  xxx2,  the  roots  have  the  same  sign. 

Since  —  p  =  —  ^  =  xl  +  x2,  the  roots  are  both  negative. 

Ex.  3.    Examine  x2  —  14  x  +  49  =  0. 

Since  p2  —  4  9  =  196  —  4  •  49  =  0,  the  roots  are  real  and  coincident. 

Ex.  4.   Examine  x2  —  7  x  +  15  =  0. 

Since  p2  —  4  5  —  49  —  4  •  15  =  —  11,  the  roots  are  imaginary. 


110  QUADRATIC  EQUATIONS 

EXERCISES 

Without  solving,  determine  the  character  of  the  roots  in  each 
of  the  following : 

1.  5a2  —  4a;- 5  =  0.  9.  16m2 +  4  =  16m. 

2.  6  x2  +  4  x  +  2  =  0.  10.  25  a2  -  10  a  =  8. 

3.  .c-'  _  4  x  +  8  =  0.  11.  20  -  13  6  —  15  62  =  0. 

4.  2  +  2  a2  =  4  a;.  12.  10  y-  +  39  y  +  14  =  0. 

5.  6  x  +  8  ar  =  9.  13.  3  a2  +  5  a  +  22. 

6.  1  -  a2  =  3  a,  14.  3  a2  -  22  a  +  21  =  0. 

7.  0  a  -  30  =  3  a2.  15.  5  b-  +66  =  27. 

8.  6a2  +  6  =  13a.  16.  Ctt-17  =  lla2. 

FORMATION   OF  EQUATIONS   WHOSE  ROOTS   ARE  GIVEN 
154.   Ex.  1.    Form  the  equation  whose  roots  are  7  and  —  4. 
From  (1)  and  (2),  §  153,  we  have 

./■j  +  x2  =  —  p  =  7  +  (  —  4)  =  3.     Hence  p  =  —  3. 
And  XjZ2  =  </  =  7(-  4)  =  -  28. 

Hence         a.-2  +  px  +  q  =  0  becomes  a;2  —  3  x  —  28  =  0. 

In  case  the  equation  is  to  have  more  than  two  roots,  we  pro- 
ceed as  in  the  following  example: 

Ex.  2.    Form  the  equation  whose  roots  are  2,  3,  and  5. 

Recalling  the  .solution  by  factoring,  we  may  write  the  desired  equa- 
tion in  the  factored  form  as  follows  : 

(x-2)(x-3)(x  -  5)  =0. 

Obviously  2,  3,  and  5,  are  the  roots  and  the  only  roots  of  this  equa- 
tion.    Hence  the  desired  equation  is: 

(x  -  2)(x  -  3)0  -  5)  =  x3  -  10  x1  +  31  x  -  30  =  0. 


EQUATIONS    WITH  GIVEN  ROOTS  111 
EXERCISES 

Form  the  equations  whose  roots  are  : 

1.  3,  -7.                    4.   5,  -4,-2.  7.    -5,  -6. 

2.  b,  c.                         5.    V5,  -  V5.  8.    -  6  +  &,  —  6  -  k. 

3.  a,  —  b,  —  c.       6.    a  —  Vo,  a  +  V3.  9.  V  —  1,  —  V—  1. 
10.   a,  -  6.             11.    8  +  V3,  8  -  V3.  12.   2,  3,  4,  5. 
13.   3  +  2  V^I,  3-2  V=T.           14.  S-V^l,  5  +  V^T. 


1C     -,    ,    ,    o  ,^     —  b  +  v'lr— 4ac    —  6  —  V&2  —  4a c 

15.     1,  -iy,  i,  o.  lb.    — ,    —  — . 

2  a  2  a 

155.  An  expression  of  the  second  degree  in  a  single  letter 
may  be  resolved  into  factors,  each  of  the  first  degree  in  that 
letter,  by  solving  a  quadratic  equation. 

Ex.  1.    Factor  6  a,-2  —  17  x  +  5. 

This  trinomial  may  be  written,  6(x2  —  ig-  x  +  |). 

Solving  the  equation,  x2  —  l7-  x  +  £  =  0,  we  find  x1  =  |  and  x2  =  §. 
Hence  by  the  factor  theorem,  §  92,  x  —  *  and  x  —  f  are  factors  of 
x2  -  J/x  +  f.     And  finally 

6(x2  —  17  x  +  5)  =  6(x  -  i)(x  -  |)  =  3(x  -  \)  ■  2(x  -  f) 
=  (%x-  l)(2x  — 5). 

This  process  is  not  needed  when  the  factors  are  rational,  but 
it  is  applicable  equally  well  when  the  factors  are  irrational  or 
imaginary. 

Ex.  2.   Factor  3<c2  +  8a>-7  =  3(ar  4-  £  a;  —  £). 

Solving  the  equation  x2  +  §  x  —  ^  =  0,  we  find, 

-  4  +  VWf        ,           -  4  -  V37 
x,  = ! and  x0  = . 

1  3  3 

Hence  as  above : 


3  x2  +  8  x 


-4  +  V37"ir        -4  -V37" 


.i[,_^±v»][._ 


-A^I-fl^bTl 


112  QUADRATIC  EQUATIONS 

EXERCISES 

In  exercises  1  to  16,  p.  110,  transpose  all  terms  of  each  equa- 
tion to  the  first  member,  and  then  factor  this  member. 

PROBLEMS    INVOLVING    QUADRATIC    EQUATIONS 

In  each  of  the  following  problems,  interpret  both  solutions 
of  the  quadratic  involved  : 

1.  The  area  of  a  rectangle  is  2400  square  feet  and  its  perim- 
eter is  200  feet.     Find  the  length  of  its  sides. 

2.  The  area  of  a  rectangle  is  a  square  feet  and  its  perimeter 
is  2  b  feet.  Find  the  length  of  its  sides.  Solve  1  by  substitu- 
tion in  the  formula  thus  obtained. 

3.  A  picture  measured  inside  the  frame  is  18  by  24  inches. 
The  area  of  the  frame  is  288  square  inches.     Find  its  width. 

4.  If  in  problem  3  the  sides  of  the  picture  are  a  and  b  and 
the  area  of  the  frame  c,  find  the  width  of  the  frame. 

5.  The  sides  a  and  b  of  a  right  triangle  are  increased  by  the 
same  amount,  thereby  increasing  the  square  on  the  hypotenuse 
by  2  Jc.     Find  by  how  much  each  side  is  increased. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stitution in  the  formula  just  obtained. 

6.  The  hypotenuse  c  and  one  side  a  are  each  increased  by 
the  same  amount,  thereby  increasing  the  square  on  the  other 
side  by  2  k.    Find  how  much  was  added  to  the  hypotenuse. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by 
substituting  in  the  formula  just  obtained. 

7.  A  rectangular  park  is  SO  by  120  rods.  Two  driveways  of 
equal  width,  one  parallel  to  the  longer  and  one  to  the  shorter 
side,  run  through  the  park.  AVhat  is  the  width  of  the  drive- 
ways it  their  combined  area  is  591  square  rods? 

8.  If  in  problem  7  the  park  is  a  rods  wide  and  b  rods  long 
and  the  area  of  the  driveways  is  c  square  rods,  find  their 
width. 


PROBLEMS  INVOLVING    QUADRATICS  113 

9.   The  diagonal  of  a  rectangle  is  a  and  its  perimeter  2  b. 
Find  its  sides. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stituting in  the  formula  just  obtained. 

10.  If  in  problem  9  the  difference  between  the  length  and 
width  is  b  and  the  diagonal  is  «,  find  the  sides.  Show  how 
one  solution  can  be  made  to  give  the  results  for  both  problems 
9  and  10. 

11.  Find  two  consecutive  integers  whose  product  is  a. 
Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by  sub- 
stituting in  the  formula  just  obtained. 

What  special  property  must  a  have  in  order  that  this  problem  may 
be  possible.     Answer  this  from  the  formula. 

12.  A  rectangular  sheet  of  tin,  12  by  16  inches,  is  made  into 
an  open  box  by  cutting  out  a  square  from  each  corner  and 
turning  up  the  sides.  Find  the  size  of  the  square  cut  out  if 
the  volume  of  the  box  is  180  cubic  inches. 

The  resulting  equation  is  of  the  third  degree.  Solve  it  by  factor- 
ing. See  §  151.  Obtain  three  results  and  determine  which  are  appli- 
cable to  the  problem. 

13.  A  square  piece  of  tin  is  made  into  an  open  box  contain- 
ing a  cubic  inches,  by  cutting  from  each  corner  a  square  whose 
side  is  b  inches  and  then  turning  up  the  sides.  Find  the 
dimensions  of  the  original  piece  of  tin. 

14.  A  rectangular  piece  of  tin  is  a  inches  longer  than  it  is  wide. 
By  cutting  from  each  corner  a  square  whose  side  is  b  inches  and 
turning  up  the  sides,  an  open  box  containing  c  cubic  inches  is 
formed.     Find  the  dimensions  of  the  original  piece  of  tin. 

15.  The  hypotenuse  of  a  right  triangle  is  20  inches  longer 
than  one  side  and  10  inches  longer  than  the  other.  Find  the 
dimensions  of  the  triangle. 

16.  If  in  problem  15  the  hypotenuse  is  a  inches  longer  than 
one  side  and  b  inches  longer  than  the  other,  find  the  dimen- 
sions of  the  triangle. 


114  QUADRATIC  EQUATIONS 

17.  The  area  of  a  circle  exceeds  that  of  a  square  by  10 
square  inches,  while  the  perimeter  of  the  circle  is  4  less  than 
that  of  the  square.  Find  the  side  of  the  square  and  the  radius 
of  the  circle. 

Use  3i  as  the  value  of  it. 

18.  If  in  problem  17  the  area  of  the  circle  exceeds  that  of 
the  square  by  a  square  inches,  while  its  perimeter  is  2  b  inches 
less  than  that  of  the  square,  find  the  dimensions  of  the  square 
and  the  circle. 

Determine  from  this  general  solution  under  what  conditions  the 
problem  is  possible. 

19.  Find  three  consecutive  integers  such  that  the  sum  of 
their  squares  is  a. 

Make  a  problem  which  is  a  special  case  of  this  and  solve  it  by 
means  of  the  formula  just  obtained.  From  the  formula  discuss  the 
cases,  a  =  2,  a  =  5,  a  =  14.  Find  another  value  of  a  for  which  the 
problem  is  possible. 

20.  The  difference  of  the  cubes  of  two  consecutive  integers 
is  397.     Find  the  integers. 

21.  The  upper  base  of  a  trapezoid  is  8  and  the  lower  base  is 
3  times  the  altitude.  Find  the  altitude  and  the  lower  base  if 
the  area  is  78. 

See  problem  7,  p.  48. 

22.  The  lower  base  of  a  trapezoid  is  4  greater  than  twice 
the  altitude,  and  the  upper  base  is  \  the  lower  base.  Find 
the  two  bases  and  the  altitude  if  the  area  is  52^-. 

23.  The  lower  base  of  a  trapezoid  is  twice  the  upper,  and  its 
area  is  72.  If  £  the  altitude  is  added  to  the  upper  base,  and 
the  lower  is  increased  by  \  of  itself,  the  area  is  then  120. 
Find  the  dimensions  of  the  trapezoid. 

24.  The  upper  base  of  a  trapezoid  is  equal  to  the  altitude, 
and  the  area  is  48.  If  the  altitude  is  decreased  by  4,  and  the 
upper  base  by  2,  the  area  is  then  14.  Find  the  dimensions  of 
the  trapezoid. 


PROBLEMS  INVOLVING   QUADRATICS  115 

25.  The  upper  base  of  a  trapezoid  is  4  more  than  4  the 
lower  base,  and  the  area  is  84.  If  the  upper  base  is  decreased 
by  5,  and  the  lower  is  increased  by  i  the  altitude,  the  area  is 
78.     Find  the  dimensions  of  the  trapezoid. 

26.  The  area  of  an  equilateral  triangle  multiplied  by  V3, 
plus  3  times  its  perimeter,  equals  81.  Find  the  side  of  the 
triangle. 

See  problem  15,  p.  236,  E.  C. 

27.  The  area  of  a  regular  hexagon  multiplied  by  V3,  minus 
twice  its  perimeter,  is  504.     Find  the  length  of  its  side. 

See  problem  20,  p.  237,  E.  C. 

28.  If  a  times  the  perimeter  of  a  regular  hexagon,  plus  V3 
times  its  area,  equals  b,  find  its  side. 

29.  The  perimeter  of  a  circle  divided  by  -rr,  plus  V3  times 
the  area  of  the  inscribed  regular  hexagon,  equals  78f.  Find 
the  radius  of  the  circle. 

30.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  plus 
the  perimeter  of  the  circle  is  a.     Find  the  radius  of  the  circle. 

31.  One  edge  of  a  rectangular  box  is  increased  6  inches, 
another  3  inches,  and  the  third  is  decreased  4  inches,  making 
a  cube  whose  volume  is  862  cubic  inches  greater  than  that  of 
the  original  box.     Find  its  dimensions. 

32.  Of  two  trains  one  runs  12  miles  per  hour  faster  than  the 
other,  and  covers  144  miles  in  one  hour  less  time.  Find  the 
speed  of  each  train. 

In  a  township  the  main  roads  run  along  the  section  lines,  one  half 
of  the  road  on  each  side  of  the  line. 

33.  Find  the  area  included  by  the  main  roads  of  a  township 
if  they  are  4  rods  wide. 

34.  If  the  area  included  by  the  main  roads  of  a  township  is 
11,190  square  rods,  find  the  width  of  the  roads. 

35.  Find  the  width  of  the  roads  in  problem  34  if  the  area 
included  by  them  is  a  square  rods. 


CHAPTER   VIII 
ALGEBRAIC  FRACTIONS 

156.  An  algebraic  fraction  is  the  indicated  quotient  of  two 

algebraic  expressions. 

ii 
Thus  —  means  n  divided  by  d. 
d  * 

From  the  definition  of  a  fraction  and  §  11,  it  follows  that 
the  product  of  a  fraction  and  its  denominator  equals  its  numerator. 

That  is,  d  •  -=  n. 

a 

REDUCTION  OF  FRACTIONS 

157.  The  form  of  a  fraction  may  be  modified  in  various  ways 
without  changing  its  value.  Any  such  transformation  is  called 
a  reduction  of  the  fraction. 

The  most  important  reductions  are  the  following : 

(A)   By  manipulation  of  signs. 

—  n  n  —  n      h  —  a  a  —  h      a  —  b 


E.g. 


d  d  —  d       —  d      c  —  d  c  —  d      d  —  c 


(B)    To  loivest  terms. 

X*  +  X2  +  1  _  (x2  +  X  +  1)(3?2  -  X  +  1) 

'9'  X6  - 1       ~  (x  -  l)(x2  +  X  +  l)(x  +  l)(x2  -  X  +  1) 

1 


(x-  l)(x  +  l) 


(C)  To  integral  or  mixed  expressions. 

2a*  +  x*  +  x  +  2     .,         1  +  -r  +  l=2z     ^x-1, 

•'  X2  +  1  X2  +  1  X2  +  1 

116 


E.g. 


REDUCTION   OF  FRACTIONS  117 

(D)   To  equivalent  fractions  having  a  common  denominator. 

.g.  — - —    and    become   respectively   ^ —     "' and 

x  +  3  z  +  2  (x  +  3)(x  +  2) 

^— -  - — *■ — ;  a  +  1  and become  respectively  a~  ~     and - 

(a;+3)(ar  +  2)  a-1  a-1  a-1 

158.  These  reductions  are  useful  in  connection  with  the 
various  operations  upon  fractions.  They  depend  upon  the 
principles  indicated  below. 

Reduction   (.4)  is  simply  an  application  of  the  law  of  signs  in 

division,  §  28.     It  is  often  needed  in  connection  with  reduction  (Z>). 

See  §  159. 

Reduction  (B)  depends  upon  the  theorem,  §  47,  —  =  -,  by  which 

bk      b 

a  common  factor  may  be  removed  from  both  terms  of  a  fraction.     It  is 

useful  in  keeping  expressions  simplified.     This  reduction  is  complete 

when  numerator  and  denominator  have  been  divided  by  their  H.  C.  F. 

See  §§  95-102. 

Reduction  (C)  is  merely  the  process  of  performing  the  indicated 
division,  the  result  being  integral  when  the  division  is  exact,  otherwise 
a  mixed  expression. 

In  case  there  is  a  remainder  after  the  division  has  been  carried 
as  far  as  possible,  this  part  of  the  quotient  can  only  be  indicated. 

Thus  »_,  +  * 

in   which   D    is   dividend,   d   is   divisor,    q   is    quotient,    and   i?   is 
remainder. 

Reduction  (D)  depends  upon  the  theorem  of  §  47,  -=  — ,  by  which 

b      kb 
a  common  factor  is  introduced  into  the  terms  of  a  fraction. 

A  fraction  is  thus  reduced  to  another  fraction  whose  denominator 
is  any  required  multiple  of  the  given  denominator. 

If  two  or  more  fractions  are  to  be  reduced  to  equivalent  fractions 
having  a  common  denominator,  this  denominator  must  be  a  common 
multiple  of  the  given  denominators,  and  for  simplicity  the  L.  C.  M.  is 
used. 


118  ALGEBRAIC  FRACTIONS 


EXERCISES 

Keduce  the  following  so  that  the  letters  in  each  factor  shall 
occur  in  alphabetical  order,  and  no  negative  sign  shall  stand 
before  a  numerator  or  denominator,  or  before  the  first  term  of 
any  factor. 

n  —  m  —  (c  —  a)  (d  —  c) 

b  —  a  (a  —  b)  (6  —  c) 

(b  —  a)(c  —  d)^  (b  —  a)(c—  b)(c  —  a) 

9.  —  ■ 


x(s—  r- 

-  (x  —  y) 

-o 

(b  —  a)  (c  —  i 
-(x-y)(z- 

—  (6  — o)(c- 

-d) 

< 

10. 


11. 


(a  —  b)  (b  —  c)  (c  —  a) 

(c  —  &  —  a)  (6  —  a  —  c) 
3  (a  —  c)  (6  —  c)  (c  —  a) 

(3c-2q)f46-a)d 


(a  —  6)  (c  —  b)(c  —  a)  ( —  a  +  6)  (a  —  &)(c  —a) 

g    _a(c  +  6)  ^     -(-r-.s)(g-Q(f-r) 

b  (c  —  a)  (n  —  m)  (—k  —  m  —  I) 


Reduce  each  of  the  following  to  lowest  terms : 

a4  -b 
a6  -  b 


13.   a4~&4,  18.     <*  +  2a?  +  2x  +  l 


14    c2-(a-&)2 

(a  +  c)2-62 


19. 


-  _       7  a#2  —  56  a^ar*  9ft 

'  28^(1 -64  a6*6)' 

16  m3  +5ffl'  +  7m  +  3  21 

///-'  +  4m  +  3 

17  a3  -  7  a  +  6  22 
a3-  7  a2  +  14a- 8*  '     cr  +  ab  +  lr  +  a  +  b 


x*  +  a 

;3_iC2_2.T_2 

2a?- 

_  ar2  _  8  a;  —  3 

2X3- 

3  or  —  7  .i'  -f  3 

4  jc8  + 

■8  a2— 3  a: +  5 

6^- 

5  .r  +  4  ,v  —  1 

a2- 

#V  +  ?/2  +  *  —  ?/  +  3 

^  +  2/ 

3  +  x2-y2  +  3x  +  3y 

a4  +  a 

2b2  +  &4  +  a3  +  '<:; 

23. 


REDUCTION  OF  FRACTIONS  119 

X-4  +  4  x?y  +  6  x2y2  +  4  an/3  +  y4  —  a4 
oj2  +  2  xy  +  y2  —  a2 


x2y  —  x2z  +  y2z  —  xy2  -f-  a-z2  —  ?/z2 

a2 — (y  + »)  &  +  yz 

25      2  a:4 -a? -20  a-2  +  16  x  -3 
3  ic4  +  5  x3  —  30  x2  —  41  a;  +  5 

3  as  _  8  a2b  -  5  air  +  6  bs 


26. 


27. 


a3  +  CM  _  9  a&2  _  9  53 

2  r3  +  r2s  +  rs2  +  2  s3 
2  ?-4  +  rs  +  3  rW  -f-  rs3  +  2  s4 


Reduce   each  of    the    following   to   an   integral   or   mixed 

expression : 

r4  4- 1  a-4  ^5 

28.   ^-X±.  30.    _^_.  32. 


a;  + 1  x  —  1  c3  +  c2  —  c  +  1 

29.    £±1.  31.    *_.         33.    ^"»  +  l 

a;  +  1  a"  +  a  +  1  ar  +  a?  + 1 

0„    a4  +  a262  +  &4  QO    ar'-ar'-a  +  l 

,54. •  00.    — — • 

a  —  0.  x3  -f-  ar  -f  a;  —  1 

3  a3  —  3  a2  +  3  a  —  1  4  m4  —  3  m3  +  3 

a  —  2  2  m2  —  2  m  +  1 

Reduce  each  of  the  following  sets  of  expressions  to  equiva- 
lent fractions  haviner  the  lowest  common  denominator: 


38. 


39. 


a;4  —  3  xry2  +  ?/4     x2  —  xy  —  y2 '    x2  +  xy—y2 

a  +  b a  b 

5 a2c  +  12 cd  —  6 ad  —  10 «o2'    5ac  —  6d'    a  — 2c 


40  a^2  +  ?/2  «  +  :>/—  x2  +  xy+j/2 


xs  +  y3  +  x2  —  xy  +  y2     x2  —  xy  +  y2       x  +  y  + 1 


120  ALGEBRAIC  FRACTIONS 

x  y 


41. 


(a  —  b)(c  —  b)(c  —  a)     (a—  b)(b  —  c)(a  —  c) 


42.    uJ5_,    _^_,    -i_,    d 
6  +  c     c  +  a     a  -f-  o 


43. 


44. 


_(6-a)(&-c)(a-c) 
6  — c  a  —  b  c  —  a 


(a—  c)(a  —  b)'    (c  — a)(6  — c)'    (6  —  a)(c  —  b) 
m  —  n  a  +  2  a  +  3 


a3  —  6  dr  + 11  a  —  6 '    a2  —  4  a  +  3 '    a2  —  3  a  4-  2 


If  o,  b,  m  are  positive  numbers,  arrange  each  of  the  follow- 
ing sets  in  decreasing  order.  Verify  the  results  by  substitut- 
ing convenient  Arabic  numbers  for  a,  b,  m. 

Suggestion.  Reduce  the  fractions  in  each  set  to  equivalent  frac- 
tions having  a  common  denominator. 

a  2  a         3a  ._         m  2  m  3  m 

45.    -,     t:,     -•         46. 


a  +  l'    a  +  2'    a  +  3  "  2m  +  l'    3m4-2'    4m  +  3 

a  4-  3  b      a  +  b      a  -\-  Ab 


47. 


a  +  4  b '    a  +  2  b '    a  +  56 


48.    Show  that,  for  a  different  from  zero,  neither  -       -   nor 
"  d  +  a 

n  ~  a  can  equal  -,  unless  n  =  d.     State  this  result  in  words, 
d  —  a  d 

and  fix  it  in  mind  as  an  impossible  reduction  of  a  fraction. 


ADDITION    AND    SUBTRACTION    OF    FRACTIONS 

159.  Fractions  which  have  a  common  denominator  are  added 
or   subtracted    in    accordance    with   the    distributive   law  for 

division,  §  30. 

rm  a  .  6      c      a  -f  b  —  c 

1  hat  is,  -  +  -  — :  =  —  —z 

d     d     d  d 

In  order  to  add  or  subtract  fractions  not  having  a  common 
denominator,  they  should  first  be  reduced  to  equivalent  frac- 
tions having  a  common  denominator. 


ADDITION   AND   SUBTRACTION   OF  FRACTIONS      121 

When  several  fractions  are  to  be  combined,  it  is  sometimes 
best  to  take  only  part  of  them  at  a  time.  In  any  case  it  is 
advantageous  to  keep  all  expressions  in  the  factored  form  as 
long  as  possible. 

Ex.      —J , i +         1 


(x  -  1)0  -  2)      (2  -  x)(x  -3)      (3  -  x)(±  -  x) 

Taking  the  first  two  together,  we  have 

1  1  2x-4  2 


(x  -  l)(s-2)      (*-2)(x-8)      (x-l)(x-2)(a!-8)      (x-l)(x-3) 

Taking  this  result  with  the  third, 

2  1  3 x- 9  3 


(x-l)(x-3)      (x-8)(x-4)      (s-l)(x-8)(x-4)       (x-l)(x-4) 

If  all  are  taken  at  once,  the  work  should  be  carried  out  as  follows : 
The  numerator  of  the  sum  is 

(x  -  3)(x  -  4)  +  (x-  1)0  -  I)  +  (x  -  1)0  -  2). 

Adding  the  first  two  terms  with  respect  to  (x  —  4),  we  have 

20-2)(z-4)  +  (x-  l)(x-2> 

Adding  these  with  respect  to  (x  —  2),  we  have  3(x  —  3)0  —  -)• 

tt          «.             •                 3(x-  3)(x  -  2) 
Hence  the  sum  is  * -^ -f- 


(x  -  l)(x-  2)0  -  3)0  -  I)      O  —  !)(*— *) 

EXERCISES 

Perform  the  following  indicated  additions  and  subtractions : 
1        2  3  4  3  5  1 


x-3     x  —  4     x-5  40  +  3)     80  +  5)     80+1) 


2  0-1)      0-2      2  0-3) 
1  7  13 


12  0  +  1)      3  0-2)      40-3) 


122 

ALGEBRAIC  FRACTIOXS 

9 

5 

,       -3,4                 „       5a; +  6           3a- -4 

(.'•  +  l)2  '  x  +  1  '  x  —  2                '    x2  -f  a:  +  1      ar  —  a;  +  1 

6             1 

x  -  2                   Q          1             4  x  -  8 

3  (x  +  1)      3  (x-  —  a;  +  1)                   .",(./•  +  2 )  '  5  ( .r  +  1 ) 

2               11 

9                                         1               ■ 

(a;  -  2)2      a;  -  2      a;  +  1 

°                    1                 r  4-  2 
10                      1                                ^        - 

(a;~2)2     6(a>-2)     5(x»  +  l) 

1\ 

1        1        1  '             * 

(as_l)«      (a,_i)      («"_!) 

12. 

1           1           3           1            3            1          1 
2  (1  -  3  xf    8  (1  -  3  a;)2  '  32  (1  -  3  x)  '  32  (1  +  a?) 

13 

1                            12 

(1  -  a)  (2  -  a)      (2  -  a) (a  -  3)   '  (3  -  a)(a  -  1) 

14 

a;  y                         yz                          xz 

(z  -  ?/) (a?  -  z)      (x-  z) (x  -  y)      (y  -  x){y  -  z) 

15 

1              2a -5          5  a2-  3  a  -  2 

a  -  1      a2  -  2  a  +  1           (a  -  l)8 

16. 

1                        1                  2m+2 

m2  +  m  +  1      nr  —  m  +  1      m4  +  »i2  +  1 

17. 

11                        2 

62  -  3  b  +  2  '  V1  -  5  6  +  6      b-  -  4  b  +  3 

18. 

r  +  s                    s  +  t                     r  +  t 

(r-t)  (s  -  0  (r  -  s)(t  -  r)      (t  -  s)  (s  -  r) 

19  p-  +  <f  |  (f-pr         (  9-°+p9 

(y>  -  7  K  P  +  '•)      (q  -  r)(q  - p)      (r  -  q)(r+p) 

20.  ;!-4-1 


5  x-  -  18  x  +  9      4  ar'  -  11  a;  -  3 


MULTIPLICATION  AND   DIVISION   OF  FRACTIONS    123 


MULTIPLICATION    AND    DIVISION    OF    FRACTIONS 

160.  Theorem.  Th  e  produ  ct  of  two  fractions  is  a  fraction 
whose  numerator  is  the  product  of  the  given  numerators 
and  whose  denominator  is  the  product  of  the  given  de- 
nominators. 

Proof.    We  are  to  prove  that  -  •  -  =  —  • 
1  b      d      bd 

Let 
Then 


X  =  - 

b 

d 

bdx  =  hi 

'a 

Kb 

d) 

§7 

bdx  =  b  ■ 

a 
b  ' 

d 

§8 

bdx  =  an. 

§H 

_  an 
X~bd 

a      n  _  an 
b  '  d~M 

§2 

Hence, 

Therefore, 

Corollary  1 .  A  fraction  is  raised  to  any  power  by  raising 
numerator  and  denominator  separately  to  that  power. 

t-i      i      ,i       i  ,i  a      a      a'2      a      a      a      a3 

ior  by  the  above  theorem,  -  •  -  =  — ,    _.-.-  =  —,  etc. 

J  b      b      b2      b      b      b      b3 

Corollary  2.  A  fraction  multiplied  by  itself  inverted 
equals  +  1. 

For  n-  ■  i  =  ^.=  +  l  and-^  •   f_^=^=  +  l. 
d      ?i      nd  d       \      nl      nd 

161.  Definitions.  If  the  product  of  two  numbers  is  +  1,  each 
is  called  the  reciprocal  of  the  other.  Hence  from  Cor.  2,  the 
reciprocal  of  a  fraction  is  the  fraction  inverted. 

Also,  since  from  ab  =  l  we  have  «  =  -  and  &  =  -,  it  follows 

b  a 

that  if  two  numbers  are  reciprocals  of  each  other,  then  either 
one  is  the  quotient  obtained  by  dividing  1  by  the  other. 


■MM 


124  ALGEBRAIC  FRACTIONS 

162.  Theorem.  To  divide  by  any  number  is  equivalent  to 
multiplying  by  its  reciprocal. 

n         1 

Proof.  We  are  to  prove  that  n  ~-  d  or  -  =  n —  This  is  an  im- 
mediate consequence  of  §  29. 

Corollary  1.  To  divide  a  n  umber  by  a  fraction  is  equiva- 
lent to  multiplying  by  the  fraction  inverted. 

For  by  §  161  the  reciprocal  of  the  fraction  is  the  fraction  inverted. 

Corollary  2.  ,  /  fraction,  is  divided  by  an  integer  by  m  7/1- 
tiph/in'J  its  denominator  or  dividing  its  numerator  by 
that  integer. 

For  -  +  a  =  -  ■  -  =  — ,  Cor.  1  and  §  160 

d  d     a      ad 

and  -  -*■  a  — ,  since  —  = by  $  4 1 . 

d  d  ad         d 

In  multiplying  and  dividing  fractions  their  terms  should  at 
once  be  put  into  factored  forms. 

When  mixed  expressions  or  sums  of  fractions  are  to  be  mul- 
tiplied or  divided,  these  operations  are  indicated  by  means  of 
parentheses,  and  the  additions  or  subtractions  within  the 
parentheses  should  be  performed  first,  §  38. 


Ex.    Simplify 


1  ,     2a2  \     f    1  1 

1  —  a  + 


1+uJ     VI  +  a      1  —  a 


«4-l 


Performing  the  indicated  operations  within  the  parentheses,  we  have 

rl  +  a2  .     2d    I        3  a8    _  1  +  a"     a2  -  1  3  a3  _      3  a2 

l-l+o  'o!-lJ'fl4-l     l  +  o'     2a     '  (V- l)(a2+l)     2(a  +  l)' 


EXERCISES 

Perform  the  following  indicated  operations  and  reduce  each 
result  to  its  simplest  form. 

1     ,r4  -f  x-y-  +  y*  m  a*2  —  f. 
Xs  —  y3  x3  +  f 


MULTIPLICATION  AND   DIVISION  OF  FRACTIONS      125 


a2  —  b'2x'2  +  ucx2  —  bcx3    _      —  ay2  —  bxy2  —  cx2y2 
30  ~a*--~Warr+  24  a  -  16  "*"  20  a;2  - 15  ao2  -  30  aV 


20  rV  +  23  rsf  —  21 12  12  wm3  -  28  mnhj  —  24  ran?/2 


8  //r'/<3  —  48  m2n2y  +  72  mr'ny*         10  rV  +  24  rs£  —  18 t2 


4.    fa      ftVa'  ,  62      a      &  ,  1N.  a5  +  6g 
\  6      aj\b2     a2      6     c<        y      a  —  6 


f . 3 w+ 2 i 
L:  2s  +  3, 


t 


5. 


my*-1 

a;     2/y\cc     W 


s-yyo  i    2-V 


*+ 


iA' 


■«  -2/ 


«  —  6      a  +  67  Ur      &-y     Va2      62 


10. 


7.      1  + 


1- 


a  —  £ 

>/i  +  n      m  —  n 

m  —  n      m  +  n 

x2  +  y2      ar  —  y' 
x2-y2      x2  +  y: 


a  +  &y     V        a2  —  62 


.       .     2ir  \     fm  +  n  .   m—n 
mi  —  »y      \m  —  ra      m  +  h 


r+r  + 


2.H/2+2.v 


x+y  +  z 


+ 


x  + 1/       (as  +  ^/)2y    V  ( •*•  +  yf—z\ 


x-  -  y 
(x  +  yf 


x+y      x—y 
v—y      x+y 


1  + 


*  +  y 


a'2  +  ab  +  b2       a  +-  b       f  ,  ,   bs  —  a2b 

11.     -  — ! •  ■ — ■  •     a  -\ 

a'2  —  ub  +  b'-      a?  —  b       V  a  +  b 


12. 


m  +  mix      m?  —  mir  —  mf  +  nr      m'-ir  +  mil3  +  rr 
m2  +  >i2  m3n  —  /i4 


13.    |ay»+afy_2^r 
x-y 


x-  +  y- 


m4  - 

2  m3  +  mr 

■*■ 

msn  +  2  m 

V2  + 

mn 

m*  - 

-u* 

1 

-I) 

or  +  y2    I 

n  _ 

31 

r 

f        { 

vr 

*yj 

126  ALGEBRAIC  FRACTIONS 

COMPLEX   FRACTIONS 

163.  A  fraction  which  contains  a  fraction  either  in  its 
numerator  or  in  its  denominator  or  in  both  is  called  a  complex 
fraction. 

Since  every  fraction  is  an  indicated  operation  in  division, 
any  complex  fraction  may  be  simplified  by  performing  the  indi- 
cated division. 

It  is  usually  better,  however,  to  remove  all  the  minor 
denominators  at  once  by  multiplying  both  terms  of  the  com- 
plex fraction  by  the  least  common  multiple  of  all  the  minor 
denominators  according  to  §  47. 


*  +  ±  ±  +  ±    .6 

I-                  i          3      2           V\      21  2x  +  3a;  ;"  /• 

r  or  example,     = =        ^        =  — _ — , 

2i_2_3      I'ljP _  3\    0       4  a:2  -  9      4  a:2  -  9 
3       2      V  3       2/ 


A  complex  fraction  may  contain  another  complex  fraction 
in  one  of  its  terms. 

E  ( , —    has  the  complex  fraction 


fl4^  «  + 


1  a  -  1 

a  + 

a  —  1 

in  its  denominator.     This  latter  fraction  is  first  reduced  by  multiply- 
ing its  numerator  and  denominator  by  a  —  1,  giving 

a  +  1  a'2  —  1 


a+      1  «»-«  +  ! 

a  —  1 


Substituting  this  result  in  the  given  fraction,  we  have 

1  _  1  _  a2  -  a  +  1 

„  + -JL±1      ~  „  +     Z2-1      "a'  +  a-l1 

1  a2  -  a  +  1 

a  + - 

a  —  1 


COMPLEX  FRACTIONS  127 


EXERCISES 

Simplify  each  of  the  following, 
m2  +  mn 


1 

m2  —  ir 

m             n 

m 

—  n      m  4  n 
a'-b4 

2. 

a2 

-2ab  +  b2 

a2  4  ab 

•A. 


a—b 

x5  —  3  x4y  +  3  xhf2  —  x2ys 

xsy  —  y4 

x>  -2  x4y  4  xAf 

xhf  4-  xf  4 1 

1  +-^-+4^-, 


a 2 

a2 

a-2+- 

a 

a2  4  1  4  -, 
a' 

a 

2      1 

-24--— 
a      or 

1 

1 

1 
1-1 

a  4  x      a  —  as      a," 


a-\-x      a  —  x      a-  —  a;2 


+  -^  +  ^1  l-1- 


a  4  *'     «>  —  x     a-  4  * 


iC 


a  —  x"      a  +  ,c      a+^ 


10. 


o              ,      q     7/13  —  n3 
■»i"  —  mr«  4  "" «  4 


■m  4  »•  o  _i_     3 


7/i"4w-4»i"H ■ —  &  + 


.»■ 


EQUATIONS    INVOLVING    ALGEBRAIC    FRACTIONS 

164.  In  solving  a  fractional  equation,  it  is  usually  conven- 
ient to  clear  it  of  fractions,  that  is,  to  transform  it  into  an  equiv- 
alent equation  containing  no  fractions. 

In  case  no  denominator  contains  any  unknown  this  may  be 
done  by  multiplying  both  members  by  the  L.  C.  M.  of  all  the 
denominators,  §  62. 


128  ALGEBRAIC  FRACTIONS 

When,  however,  the  unknown  appears  in  any  denominator, 
multiplying  by  the  L.  C.  M.  of  all  the  denominators  may  or  may 
not  introduce  new  roots,  as  shown  in  the  following  examples. 

It  may  easily  be  shown,  that  multiplying  an  integral  equation 
by  any  expression  containing  the  unknown  always  introduces  new 
roots. 

Ex.1.    Solve  JL_  +  ^!_  =  2.  (1) 

x-2     x-'S  v  ' 

Clearing  of  fractions  by  multiplying  by  (x  —  2)(.r  —  3),  and  simpli- 
fying, we  have     .    „      ,_         _rt      ,        lwo         _. 

J     &'  2  ./ -  -  13  x  +  20  =  (x  -  4)  (2  x  -  5)  =  0.  (2) 

The  roots  of  (2)  are  4  and  21,,  both  of  which  satisfy  (1).  Hence 
no  new  root  was  introduced  by  clearing  of  fractions. 

Ex.2.    Solve  — 1—  = .  (1) 

jc_1      (a;-l)(a;_2) 

Clearing  of  fractions,  we  have, 

a; -2  =  1.  (2) 

The  only  root  of  (2)  is  x  =  3,  which  is  also  the  only  root  of  (1). 
Hence  no  new  root  was  introduced. 

Ex.  3.    Solve  ^fll  =  i.  (1) 

x2  -  4 

Clearing  of  fractions  and  simplifying,  we  have, 

a!2_a;_2  =  (a;_2)(s  +  l)  =  0.  (2) 

The  roots  of  (2)  are  2  and  —  1.  Now  x  =  —  1  is  a  root  of  (1).  but 
x  =  2  is  not,  since  we  are  not  permitted  to  make  a  substitution  which 
reduces  a  denominator  to  zero,  §  50.  Hence  a  new  root  has  been  in- 
troduce'! and  (1)  and  (2)  are  not  equivalent. 

x  -  2 

If  the  fraction  ~r, r  is  first  reduced  to  lowest  terms,  we  have  the 

x2  —  4 

eiiuation  -i 

— —  =  1.  (3) 

x  +  2  K  } 

Clearing  of  fractions,  x  +  2  =  1.  (4) 

Now  (:'>)  and  (  I)  are  equivalent,  —  1  being  the  only  root  of  each. 


EQUATIONS  INVOLVING   FRACTIONS  129 

Ex.4.    Solve  _i^__^±l  =  i.  (i) 

x-  —  1       X  —  1 

dealing  of  fractions  and  simplifying, 

x'2  -  x  =  x(x  -  1)  =  0.  (2) 

The  roots  of  (2)  are  0  and  1.  x  =  0  satisfies  (1),  but  x  =  1  does 
not,  since  it  is  not  a  permissible  substitution  in  either  fraction  of  (1). 
Hence  a  new  root  has  been  introduced. 

165.  Examples  3  and  4  illustrate  the  only  cases  in  which  new 
roots  can  be  introduced  by  multiplying  by  the  L.  C.  M.  of  the 
denominators. 

This  can  be  shown  by  proving  certain  important  theorems, 
the  results  of  which  are  here  used  in  the  following  directions 
for  solving  fractional  equations  : 

(1)  Reduce  all  fractions  to  their  lowest  terms. 

(2)  Multiply  both  members  by  the  least  common  multiple 
of  the  denominators. 

(3)  Reject  any  root  of  the  resulting  equation  which  reduces 
any  denominator  of  the  given  equation  to  zero.  The  remain- 
ing roots  will  then  satisfy  both  equations,  and  hence  are  the 
solutions  desired. 

If  when  each  fraction  is  in  its  lowest  terms  the  given  equation  con- 
tains no  two  which  have  a  factor  common  to  their  denominators,  then 
no  new  root  can  enter  the  resulting  equation  and  none  need  to  be  re- 
jected.    See  Ex.  1  and  Ex.  3  after  being  reduced. 

If,  however,  any  two  or  more  denominators  have  some  common 
factor  x  —  a,  then  x  =  a  may  or  may  not  be  a  new  root  in  the  resulting 
equation,  but  in  any  case  it  is  the  only  possible  kind  of  new  root 
which  can  enter,  and  must  be  tested.     Compare  Exs.  2  and  4. 

Ex.5.    Examine        3x  +  7      + — ^±1  =  8. 

3^  +  2  as +  11      a?  +  3x  +  2      x  —  1 

Since  each  fraction  is  in  its  lowest  terms  and  no  two  denominators 
contain  a  common  factor,  then  clearing  of  fractions  will  give  an  equa- 
tion equivalent  to  the  given  one. 


130  ALGEBRAIC  FRACTIONS 

Lx.  6.    Examine  — ! —  —  =  4. 

ar  +  ox  +  G      xr  +  3  x  4-  * 

Each  fraction  is  in  its  lowest  terms,  but  the  two  denominators  have 
the  factor  x  +  2  in  common.  Hence  x  =  —  2  is  the  only  possible  new- 
root  which  can  enter  the  resulting  integral  equation,  but  on  trial  it  is 
fou ml  not  to  be  a  root.     Hence  the  two  equations  are  equivalent. 

EXERCISES 

Determine  whether  each  of  the  following  when  cleared  of 
fractions  produces  an  equivalent  equation,  and  solve  each. 

2.    ±±**±±  =  2x  +  S. 


3  x2  —  7  x  +  3  x-  —  4 

3  3 |  °  |  1        _Q 

2^-a-l      f-1      s+1 

.                —  t C           .            Ju                  O  it  -* 

4. r- =    —  1. 

2  a  -  1      a-  +  1      a  -  1 


1                .        .,,       a  +  x        ,9      6  +  x 
u-b ' —  =  air — 


3(aj-l)      a,-2-l      4  b 

6.    2a-l  +  l=     3a    .  1Q     ^^hi, 
a          2     3«  — 1  1  — okb 

2  3  fi  " 


.;■  —  a      .r  —  6      .t  —  c                     .r  —  40  10  —  .'• 

1  1  .">  +  x-  (')  —  .<•      .r  —  4 c 


a  —  x     a  4-  .'■  a2  —  or  cc  —  4      6  —  a;      d 

13  «        |        24      =  2(«-4)      1. 

2a-l      4a--l       2  a +  1       9 

a         a  —  1  _  a2  +  a  —  1 

a  —  \  a  a2  —  a 

15.    -J 2_  =  i  +  _J 

a2  — 4      2  — a  3(a  +  2) 


EQUATIONS  INVOLVING   FRACTIONS  131 

ax  +  b  .  ox  +  d      ax  —  b  .  ex  —  d 

16.  1 = 1 • 

a  +  bx     e  +  dx      a  —  bx      e  —  dx 

(a  —  x)(x  —  b)     _  ,  _      x  +  m  —  2  n.  _  n  -4-  2  m  —  2  a-  _ 

(a  —  .r)  —  (a;  —  6)  x  +  ?/t  -f  2  n      n  —  2  m  4-  2  .« 

(a  —  xf  —  (x—  b)2  _    4  afr 
(a  —  x)  (x  —  b)         a2  —  b2 

20     1+3x     9~lla;  =  ll      (2a-~3y 
5  4- 7  a;      5  —  7x  25  —  49  x2 

x  +  2a  ,x  —  2a  _     4a& 


2  6  — oj     2  6  +  a;     4  ft2  — a;2 

22.  JL+        7x        =-J— 

.r  -2      24  (as  +  2)      a~  -  4 

23.  3?  + ft  (  a;-«=      2  (a2  +  1) 

a;  —  a      a:  +  a      (1  +  «)  (1  —  a) 

24.  *  =  *  =  »=£.  25.     -1 **-  =  b. 

x  +  hi      n  +  .f  a;  —  a      x1  —  a2 

26.    _4_  +  4(3*-l)  =  3j?L±l. 

3  a;  + 1        2  a;  + 1        3  x  +  1 

2s  +  3    +  T-3s        s-7    =Q 
2(2  x  - 1)      3  x-  4      2(a;  +  1) 

28.     J_  +  «_^  =  _J_+£±^. 
a  —  6         a;         a  +  b         x 

1  p 

29 


3  (m  +  » ■)"      w  +  »      2  (m  +  n) 
p2x  p 

30    -/2  + 2  ?/  ~ 2  +   y   _.   ?/   . 
y*  +  5y  +  6     ?/  +  3    y  +  2* 


132  ALGEBRAIC  FRACTIONS 

__  5  7       _  8  a:2  —  13  a;  —  64 

2  a; +  3      3  a;  —  4         6  ar*  +  sc  —  12 


«•  ^'-,-^- 


33. 


sc3  —  a8     a;  —  a      x2  +  ckb  +  a? 

1  _  l>  x     5  _  6  a;     8  1  -  3  a8 


3  -  4  x-     7  —  8  aj     3     21  -  52  cc  +  32  a~ 


34.     m~9  |   n-'p  =  m-q      n-p 
x  —  n      x  —  q      x  —  p     x  —  in 


35.  -* ^-  +  -        -5-       -  =  0. 

aj-3      a; -2      4.^ -20a; +  24 

or  q  i 

36.  — ; +  _i_  =  0. 

.c:i  +  27      x-  —  3  x  +  9      as  +  3 


37. 


.c 


9      .'■  —  7      s»  —  9      &•  —  8      x  —  7     x  — 


-  5      x  —  2      x  —  4      a-  —  5      x  —  4      a;  —  2 

38      q  —  ^  +  h  -  c)  (x  ~  h  +  c)  . 
(6  +  c  +  x)  (6  +  c  —  x) 

PROBLEMS 

1.  Find  a  number  such  that  if  it  is  added  to  each  term  of 
the  fraction  f  and  subtracted  from  each  term  of  the  fraction 
i|  the  results  will  be  equal. 

2.  Make  and  solve  a  general  problem  of  which  1  is  a 
special  case. 

3.  Three  times  one  of  two  numbers  is  4  times  the  other. 
If  the  sum  of  their  squares  is  divided  by  the  sum  of  the  num- 
bers, the  quotient  is  42$  less  than  that  obtained  by  dividing 
the  sum  of  the  squares  by  the  difference  of  the  numbers.  Find 
the  numbers. 


PROBLEMS  133 

4.  The  sum  of  two  numbers  less  2,  divided  by  their  differ- 
ence, is  4,  and  the  sum  of  their  cubes  divided  by  the  difference 
of  their  squares  is  If  times  their  sum.     Find  the  numbers. 

5.  The  circumference  of  the  rear  wheel  of  a  carriage  is 
4  feet  greater  than  that  of  the  front  wheel.  In  running  one 
mile  the  front  wheel  makes  110  revolutions  more  than  the  rear 
wheel.     Find  the  circumference  of  each  wheel. 

6.  State  and  solve  a  general  problem  of  which  5  is  a  special 
case,  using  b  feet  instead  of  one  mile,  letting  the  other  num- 
bers remain  as  they  are  in  problem  5. 

7.  In  going  one  mile  the  front  wheel  of  a  carriage  makes 
88  revolutions  more  than  the  rear  wheel.  If  one  foot  is  added 
to  the  circumference  of  the  rear  wheel,  and  3  feet  to  that  of 
the  front  wheel,  the  latter  will  make  22  revolutions  more  than 
the  former.     Find  the  circumference  of  each  wheel. 

8.  State  and  solve  a  general  problem  of  which  7  is  a  special 
case,  using  a  instead  of  88,  letting  the  other  numbers  remain 
as  they  are. 

9.  The  circumference  of  the  front  wheel  of  a  carriage  is 
a  feet,  and  that  of  the  rear  wheel  b  feet.  In  going  a  certain 
distance  the  front  wheel  makes  n  revolutions  more  than  the 
rear  wheel.     Find  the  distance. 

10.  State  and  solve  a  problem  which  is  a  special  case  of 
problem  9,  using  the  formula  just  obtained. 

11.  There  is  a  number  consisting  of  two  digits  whose  sum, 
divided  by  their  difference,  is  4.  The  number  divided  by  the 
sum  of  its  digits  is  equal  to  twice  the  digit  in  units'  place 
plus  -|-  of  the  digit  in  tens'  place.     Find  the  number. 

12.  There  is  a  fraction  such  that  if  3  is  added  to  each  of  its 
terms,  the  result  is  4,  and  if  3  is  subtracted  from  each  of  its 
terms,  the  result  is  ^.     Find  the  fraction. 

13.  State  and  solve  a  general  problem  of  which  12  is  a 
special  case. 


134  ALGEBRAIC  FRACTIONS 

14.  A  and  B  working  together  can  do  a  piece  of  work  in 
6  days.  A  can  do  it  alone  in  5  days  less  than  B.  J  low  long 
will  it  require  each  when  working  alone'.' 

15.  State  and  solve  a  general  problem  cf  which  14  is  a 
special  case. 

16.  On  her  second  westward  trip  the  Mauritania  traveled 
625  knots  in  a  certain  time.  If  her  speed  had  been  5  knots 
less  per  hour,  it  would  have  required  6\  hours  longer  to  cover 
the  same  distance.     Find  her  speed  per  hour. 

17.  By  increasing  the  speed  a  miles  per  hour,  it  requires 
b  hours  less  to  go  c  miles.  Find  the  original  speed.  Show 
how  problem  16  may  be  solved  by  means  of  the  formula  thus 
obtained. 

18.  A  train  is  to  run  d  miles  in  a  hours.  After  going  c  miles 
a  dispatch  is  received  requiring  the  train  to  reach  its  destina- 
tion b  hours  earlier.  What  must  be  the  speed  of  the  train  for 
the  remainder  of  the  journey  ? 

19.  A  man  can  row  a  miles  down  stream  and  return  in  b 
hours.  If  his  rate  up  stream  is  c  miles  per  hour  less  than 
down  stream,  find  the  rate  of  the  current,  and  the  rate  of  the 
boat  in  still  water. 

20.  State  and  solve  a  special  case  of  problem  19. 

21.  A  can  do  a  piece  of  work  in  a  days,  B  can  do  it  in  b  days, 
and  (1  in  c  days.  How  long  will  it  require  all  working  to- 
gether to  do  it  ? 

22.  Three  partners,  A,  B,  and  C,  are  to  divide  a  profit  of  p 
dollars.  A  had  put  in  a  dollars  for  m  months,  B  had  put  in 
b  dollars  for  //  months,  and  C  c  dollars  for  t  months.  What 
share  of  the  profit  does  each  get? 

23.  State  and  solve  a  problem  which  is  a  special  case  of  the 
preceding  problem. 


CHAPTER   IX 

RATIO,   VARIATION,   AND   PROPORTION 

RATIO   AND   VARIATION 

166.  In  many  important  applications  fractions  are  called 
ratios. 

E.g.    f  is  called  the  ratio  of  3  to  5  and  is  sometimes  written  3  :  5. 

It  is  to  be  understood  that  a  ratio  is  the  quotient  of  two  num- 
bers and  hence  is  itself  a  number.  We  sometimes  speak  of  the 
ratio  of  two  magnitudes  of  the  same  kind,  meaning  thereby 
that  these  magnitudes  are  expressed  in  terms  of  a  common  unit 
and  a  ratio  formed  from  the  resulting   numbers. 

E.g.  If,  on  measuring,  the  heights  of  two  trees  are  found  to  be  25 
feet  and  35  feet  respectively,  we  say  the  ratio  of  their  heights  is  ff  or  f. 

167.  Two  magnitudes  are  said  to  be  incommensurable  if  there 
is  no  common  unit  of  measure  which  is  contained  exactly  an 
integral  number  of  times  in  each. 

E.g.     If  a  and  d  are  the  lengths  of  the  side  and  the  diagonal  of  a 

square,  then  d'2  —  a'2  +  a'2.  §  151,  E.  C.     Hence,  — =  -  or  -  =  — -.    But 

d2     '2        d      -y/2 

since   V'2  is  neither  an  integer  nor  a  fraction  (§  108),  it  follows  that  a 
and  d  have  uo  common  measure,  that  is,  they  are  incommensurable. 

168.  In  many  problems,  especially  in  Physics,  magnitudes 
are  considered  which  are  constantly  changing.  Number  ex- 
pressions representing  such  magnitudes  are  called  variables, 
Avhile  those  which  represent  fixed  magnitudes  are  constants. 

E.g.  Suppose  a  body  is  moving  at  a  uniform  rate  of  5  ft.  per 
second.  If  t  is  the  number  of  seconds  from  the  time  of  starting 
and  s  the  number  of  feet  passed  over,  then  s  and  t  are  variables. 

135 


13(3  RATIO,    VARIATION^   AND  PROPORTION 

The  variables  s  and  t,  in  case  of  uriform  motion,  have  a.  fixed  ratio; 
namely,  in  this  example,  s:t  =  5  for  every  pair  of  corresponding 
values  of  s  and  t  throughout  the  period  of  motion. 

169.  When  two  variables  are  so  related  that  for  all  pairs  of 
corresponding  values,  their  ratio  remains  constant,  then  each 
one  is  said  to  vary  directly  as  the  other. 

E.g.  If  s  :  t  =  k  (a  constant)  then  s  varies  directly  as  t,  and  t  varies 
directly  as  s. 

Variation  is  sometimes  indicated  by  the  symbol  oc.     Thus 

sac  t  means  s  varies  as  t,  i.e.  -  =  k  or  s  =  kt. 

t 

170.  When  two  variables  are  so  related  that  for  all  pairs  of 
corresponding  values  their  product  remains  constant,  than  each 
one  is  said  to  vary  inversely  as  the  other. 

E.g.  Consider  a  rectangle  whose  area  is  A  and  whose  base  and 
altitude  are  b  and  h  respectively.     Then,  A  =  h  •  b. 

If  now  the  base  is  multiplied  by  2,  3,  4,  etc.,  while  the  altitude  is 
divided  by  2,  3,  4,  etc.,  then  the  area  will  remain  constant.  Hence, 
b  and  h  may  both  run/  while  .4  remains  constant. 

The  relation  b  ■  h  =  A   may  be  written  b  =  A  •  -  or  h  —  A  ■  -•     It 

1  1  h  b 

mav  also  be  written  b  :  —  =  A    or  h:  ~  —  A,  so  that  the  ratio  of  either 

h  b 

b  or  /<  to  the  reciprocal  of  the  other  is  the  constant  A.     For  this  rea- 
son one  is  said  to  vnr;/  i/tn  rsr///  as  tit?  other. 

171.  If  y  =  A.r,  k  being  constant  and  x  and  y  variables,  then 

y  varies  directly  as  x".     If  y  =  — ,  then  y  varies  inversely  as  x2. 

xr 
If  y  =  k  ■  ivx,  then  y  varies  jointly  as  w  and  x.     If  y  oc  wx,  then  y 

x  w  if  x  is  constant  and  y  cc  x  if  w  is  constant.    If  y  =  A-  •  — ,  then 

a; 
y  varies  directly  as  w  and  inversely  as  x. 

Example.  The  resistance  offered  by  a  wire  to  an  electric 
current  varies  directly  as  its  length  and  inversely  as  the  area 
of  its  cross  section. 

If  a  wire  |  in.  in  diameter  has  a  resistance  of  r  units  per  mile, 
find  the  resistance  of  a  wire  \  in.  in  diameter  and  3  miles  long. 


RATIO  AND    VARIATION  137 

Solution.  Let  R  represent  the  resistance  of  a  wire  of  length  I  and 
cross-section  area  s  =  tt  •  (radius)2.  Then  R  =  k  ■  -  where  k  is  some 
constant.         Since    R  =  r    when   1  =  1    and    s  =  ^(y^)2,    we    have 

1  ,  77T 

r  —  I- . or  k  = 

r  ~  k    _7r_  256 

256 
Hence,  when  I  =  3  and  s  =  7r(|)2,  we  have, 

j,  _   TIT       3    _  3 

256  '  jr       4  ?  ' 
64 

That  is,  the  resistance  of  three  miles  of  the  second  wire  is 
|  the  resistance  per  mile  of  the  first  wire. 

PROBLEMS 

1.  If  zee  id,  and  if  z  =  27  when  w  =  3,  find  the  value  of  z 
when  w  =  4^. 

2.  If  2  varies  jointly  as  to  and  a*,  and  if  z  =  24  when  ro  =  2 
and  x  =  3,  find  z  when  w  =  '3\  and  x  =  7. 

3.  If  z  varies  inversely  as  w,  and  if  2  =  11  when  w  =  3, 
find  2  when  w  =  G6. 

4.  If  z  varies  directly  as  to  and  inversely  as  x,  and  if  z  =  28 
when  zc  =  14  and  x  =  2,  find  z  when  w  =  42  and  x  =  3. 

5.  If  2  varies  inversely  as  the  square  of  w,  and  if  z  =  3 
when  ^c  =  2,  find  z  when  w  =  6. 

6.  If  7  varies  directly  as  m  and  inversely  as  the  square 
of  rZ,  and  q  =  30  when  m  —  1  and  d  =  T^,  find  q  when  m  =  3 
and  d  =  GOO. 

7.  If  ?/2  oc  xs,  and  if  ^/  =  16  when  x  =  4,  find  ?/  when  a*  =  9. 

8.  The  weight  of  a  triangle  cut  from  a  steel  plate  of  uni- 
form thickness  varies  jointly  as  its  base  and  altitude.  Find 
the  base  when  the  altitude  is  4  and  the  weight  72,  if  it  is 
known  that  the  weight  is  00  when  the  altitude  is  5  and  base  6. 


138  RATIO,    VARIATION,   AND  PROPORTION 

9.  The  weight  of  a  circular  piece  of  steel  cut  from  a  sheet 
of  uniform  thickness  varies  as  the  square  of  its  radius.  Find 
the  weight  of  a  piece  whose  radius  is  13  ft.,  if  a  piece  of 
radius  7  feet  weighs  196  pounds. 

10.  If  a  body  starts  falling  from  rest,  its  velocity  varies 
directly  as  the  number  of  seconds  during  which  it  has  fallen. 
If  the  velocity  at  the  end  of  3  seconds  is  96.6  feet  per  second, 
find  its  velocity  at  the  end  of  7  seconds ;  of  ten  seconds. 

11.  If  a  body  starts  falling  from  rest,  the  total  distance 
fallen  varies  directly  as  the  square  of  the  time  during  which  it 
has  fallen.  If  in  2  seconds  it  falls  64.4  feet,  how  far  will  it 
fall  in  5  seconds  ?     In  9  seconds  ? 

12.  The  number  of  vibrations  per  second  of  a  pendulum 
varies  inversely  as  the  square  root  of  the  length.  If  a  pen- 
dulum 39.1  inches  long  vibrates  once  in  each  second,  how 
long  is  a  pendulum  which  vibrates  3  times  in  each  second  ? 

13.  Illuminating  gas  in  cities  is  forced  through  the  pipes  by 
subjecting  it  to  pressure  in  the  storage  tanks.  It  is  found  that 
the  volume  of  gas  varies  inversely  as  the  pressure.  A  certain 
body  of  gas  occupies  49,000  cu.  ft.  when  under  a  pressure  of  2 
pounds  per  square  inch.  What  space  would  it  occupy  under 
a  pressure  of  2i  pounds  per  square  inch  ? 

14.  The  amount  of  heat  received  from  a  stove  varies  in- 
versely as  the  square  of  the  distance  from  it.  A  person  sitting 
15  feet  from  the  stove  moves  up  to  5  feet  from  it.  How  much 
will  this  increase  the  amount  of  heat  received? 

15.  The  weights  of  bodies  of  the  same  shape  and  of  the  same 
material  vary  as  the  cubes  of  corresponding  dimensions.  If  a 
ball  3*  inches  in  diameter  weighs  14  oz.,  how  much  will  a  ball 
of  the  same  material  weigh  whose  diameter  is  3.1,  inches? 

16.  On  the  principle  of  problem  15,  if  a  man  5  feet  9  inches 
tall  weighs  165  pounds,  what  should  be  the  weight  of  a  man  of 
similar  build  6  feet  tall  ? 


PROPORTION  139 

PROPORTION 

172.  Definitions.  The  four  numbers  a,  b,  c,  d  are  said  to  be 
proportional  or  to  form  a  proportion  if  the  ratio  of  a  to  b  is  equal 

to  the  ratio  of  c  to  d.     That  is,  if  -=-•    This  is  also  sometimes 

'      b     d 

written  a:b::c:d,  and  is  read  a  is  to  b  as  c  is  to  d. 

The  four  numbers  are  called  the  terms  of  the  proportion ; 
the  first  and  fourth  are  the  extremes;  the  second  and  third  the 
means  of  the  proportion.  The  first  and  third  are  the  antece- 
dents of  the  ratios,  the  second  and  fourth  the  consequents. 

If  a,  b,  c,  x  are  proportional,  x  is  called  the  fourth  propor- 
tional to  a,  b,  c.  If  a,  x,  x,  b  are  proportional,  x  is  called  a  mean 
proportional  to  a  and  b,  and  b  a  third  proportional  to  a  and  x. 

173.  If  four  numbers  are  proportional  when  taken  in  a  given 
order,  there  are  other  orders  in  which  they  are  also  proportional. 

E.g.  If  ii,  b,  c,  d  are  proportional  in  this  order,  they  are  also  pro- 
portional in  the  following  orders:  a,  c,  b,  d;  b,  a,  d,  c;  b,  d,  a,  c; 
c,  a,  d,  b ;  c,  d,  a,  b ;  d,  c,  b,  a  :  and  d,  b,  c,  a. 

Ex.  1.  Write  in  the  form  of  an  equation  the  proportion  cor- 
responding to  each  set  of  four  numbers  given  above,  and  show 

how  each  may  be  derived  from  -  =  —     See  §  196,  E.  C. 

b      d 

Show  first  how  to  derive  -  =  -  (1),  and  then  -  =  -  (2). 
c      d  a     c 

t\    •        i  a  +  b     c  +  d  /0  v         ,a  —  6      c  —  d  .  ,s 

Derive  also        — ! —  = (.3),  and  = (1). 

a  c  a  c 

In  (1)  the  original  proportion  is  said  to  be  taken  by  alterna- 
tion, and  in  (2)  by  inversion ;  in  (3)  by  composition,  and  in  (4) 
by  division. 

Ex.  2.    From  -  =  -  and  (1),  (2),  (3),  (4)  obtain  the  following. 

See  pp.  279-281,  E.  C. 

a±b _c ±d  a  ±b  _a  a±c  _a  a  ±b  _a  —  b 

b  d  c  ±d      c '  b±d      b'  c  +  d      cTd' 

When  the  double  sign  occurs,  the  upper  signs  are  to  be  read  together 
and  the  lower  si<nis  together. 


X 
■YZ^Z}.  MCALS 

•:r  won!  exponent 

■     -  - 

■ 

------- 

/  rjfguarmeum»  aAikk  gov- 

-  .- 

- 

V  :"  =  : 

V  -  V  =  v"-' 

0       '    =   C'n"  -'-'-' 

%  116 

- 

•■•  -     - 

-rii 

■ 


FBACTIONA        VD   NEGATIVE  EXPONENTS       143 

/I'll  r 

Similarly,  from    \b'      t?  •  ft*.  ..to  r  factoi   .       l>% 
we  show  that  ■    \b"  J  =  (i'h  )  . 

Hence,  =  vV  =  (Vft  ) '.  See  ;'  119 

Thus  a  posit  i-  ina]  exponent  means  a  roo£  0/  a  power 

or  a  power  of  a  roi      e  numerator  indicating  the  power  and  the 
denominator  indica      the  root. 

E.g.    J=  Vol  =  ■  </  )-':  v       v  •;  J        I,  or  (v^)2  =  22  =  4. 

177.  Assuming  I  I  to  hold  also  for  negative  exponents, 
and  letting  £  be  a  tive  number,  integral  or  fractional,  we 
determine  as  follow  the  meaning  of  6  '  (read  b  exponent  nega- 
tive t). 

By  LawT,  V  ■  b~l  =  ft°  =  1.  .:  16 

Therefore,  ft-'  =  -•  §11 

ft' 

Hence  a  numberwith  a  negative  exponent  means  the  same 
as  the  reciprocal  qjthe  number  with,  a  positive  exponent  of  the 

same  absolute  i 

E.q.     a"2=-:       L"*=i-  =  !  =  1. 

a8'  4|      23      8 

178.  It  thus  appars  that  fractional  and  negative  exponents 
simply  provide  n  ways  of  indicating  operations  already  well 
known.     Sometime  one  notation  is  more  convenient  and  s< 

times  the  other. 

Fractional  and  cgative  exponents  are  also  called  powers. 

E.g.  x?  may  be  3ad  x  to  the  f  power,  and  x~*  may  be  read  x  to  the 
—  4  th  power. 

The   limitation-    a   to   principal  roots   and   tl  i   the   base, 

imposed  in   theoi     •   on   powers    and    roots   in    Chapter  VT, 
sarily    ap]  lv   to  cot  responding   theorems   in    this  chapter.      S 

§§  114-122." 


CHAPTER   X 
EXPONENTS   AND   RADICALS 

FRACTIONAL    AND    NEGATIVE    EXPONENTS 

174.  The  meaning  heretofore  attached  to  the  word  exponent 
cannot  apply  to  a  fractional  or  negative  number. 

E.g.  Such  an  exponent  as  §  or  —  5  cannot  indicate  the  number  of 
times  a  base  is  used  as  a  factor. 

It  is  possible,  however,  to  interpret  fractional  and  negative 
exponents  in  such  a  way  that  the  laws  of  operations  which  gov- 
ern positive  integral  exponents  shall  apply  to  these  also. 

175.  The  laws  for  positive  integral  exponents  are : 

I.  a'"  ■  a"  =  am+n.  §  43 

II.  a'"  h-  a"  =  a"'-".  §  46 

III.  (a'")"  =  a"'".  §  115 

IV.  (a'"  ■  b")p  =  a"'Pb"P.  §  116 

V.  (a'"  -f-  s")p  =  a'"P  -7-  s"P.  §  117 

176.  Assuming  Law  I  to  hold  for  positive  fractional  expo- 
nents and  letting  /•  and  s  be  positive  integers,  we  determine  as 

follows  the  meaning  of  1/  (read  l>  exponent  r  divided  by  s). 

, ,    .     ( -Y    -  - 

By  definition,  \lf  J  =  Ir  ■  Ir  ■••  to  s  factors, 

r      r  r 

which  by  Law  I  =  h*    •     ■■■  to  s  terms  =  6s     =  br. 

Hence.  Ir  is  one  of  I  lie  s  equal  factors  of  lr. 
r  l 

That  is,  b*  -  \ 7/  .  and  in  parfcictHar  6*  =  VB .  See  §  114 

142 


FRACTIONAL  AND  NEGATIVE  EXPONENTS        143 

/  IV      I      1  = 

Similarly,  from   \b"  J  =  b*  ■  bs  •••to  r  factors,  =  b", 

we  show  that        .  6s  =  \ff )  =  (vV)  . 

Hence,  6*  =  </F  =  (VFY.  See  §  119 

Thus  a  positive  fractional  exponent  means  a  root  of  a  power 
or  a  power  of  a  root,  the  numerator  indicating  the  power  and  the 
denominator  indicating  the  root. 

E.g.     J  =  Iffi  =  (V^)2;  8*=  ^6?  =  4,  or  (v^)*=  22  =  4. 

177.  Assuming  Law  I  to  hold  also  for  negative  exponents, 
and  letting  t  be  a  positive  number,  integral  or  fractional,  we 
determine  as  follows  the  meaning  of  b~*  (read  b  exponent  nega- 
tive t). 

By  Law  I,  V  ■  Jr*  =  b°  =  1.  §46 

Therefore,  b~<  =  -■  §  11 

Hence  a  number  with  a  negative  exponent  means  the  same 
as  the  reciprocal  of  the  number  with  a  positive  exponent  of  the 
same  absolute  value. 

'9'     a  a2'  4f      28      8* 

178.  It  thus  appears  that  fractional  and  negative  exponents 
simply  provide  new  ways  of  indicating  operations  already  well 
known.  Sometimes  one  notation  is  more  convenient  and  some- 
times the  other. 

Fractional  and  negative  exponents  are  also  called  powers. 

2 

E.g.  xs  may  be  read  x  to  the  j  power,  and  x~4  may  be  read  x  to  the 
—  -ith  -power. 

The  limitations  as  to  principal  roots  and  the  sign  of  the  base, 
imposed  in  theorems  on  powers  and  roots  in  Chapter  VI,  neces- 
sarily apply  to  the  corresponding  theorems  in  this  chapter.  See 
§§  114-122. 


144  EXPONENTS  AND   RADICALS 

In  any  algebraic  expression,  radical  signs  may  now  be  re- 
placed by  fractional  exponents,  or  fractional  exponents  by 
radical  signs. 

In  a  fraction,  any  factor  may  be  changed  from  numerator  to 
denominator,  or  from  denominator  to  numerator,  by  changing 
the  sign  of  its  exponent. 

Z/—T  5/ I—  7/ 3/ —  %  3      '  4      2 

Ex.  1.    vr  +  3  vx3  •  Vi/  +  5  w  v y-  =  x3  -f  8zTy*  +  ox'y*. 
r.x.  2.    —  =  ata  -,  siuce  aox  z  =  ab  •  —  =  —  • 


r<        o         t-a  5              •>       1         '"'" 
Lx.  •>.    alr6c-  =  «c-  •  —  = 

¥        63 

-4       1            1 
Ex.  4.    32  5  =  -±-  =  — i = 

82*      (^32> 

1  _  1 
24  ~  16 

EXERCISES 

(n)  In  the  expressions  containing  radicals  on  p.  151,  replace  these 
by  fractional  exponents. 

(b)  Replace  all  positive  fractional  exponents  on  this  page  by 
radicals. 

(<■)  Change  all  expressions  containing  negative  exponents  to  equiv- 
alent expressions  having  only  positive  exponents. 

179.  Fractional  and  negative  exponents  have  been  defined 
so  as  to  conform  to  Law  I,  §§  176,  177.  We  now  show  thai 
when  so  defined  they  also  conform  to  Laws  II,  III,  IV,  and  V. 

To  verify  Law  II.  Since  by  Law  I.  nm-"  ■  nn  =  nm,  for  m  and  n  inte- 
gral or  fractional,  positive  or  negative,  it  follows  by  §  11  that 
am  •*-  an  —  am~n  for  all  rational  exponents. 

To  verify  Law  III.  Let  r  and  .<  be  positive  integers,  and  let  k  be 
any  positive  or  negative  integer  or  fraction.     Then  we  have  : 

r  r*  r 

(1)  (f,*y  =  </(a*y'  =  Vci*?  =  a  '  =  a'     ,  by  §§  170,  115. 

(2)  (a*)"  =— L_  =  _J_  =  «"V*.  by  §  177  and  (1). 

(aky     a' ' * 
Hence  (a*)"  =  a"1'  for  all  rational  values  of  n  and  k. 


FRACTIONAL   AND   NEGATIVE  EXPONENTS        145 

To  verify  Law  IV.  Let  m  and  n  be  positive  or  negative  integers  or 
fractions,  and  let  r  and  s  be  positive  integers,  then  we  have 

(1)  («W/')'  =  </'(('"'/>")'■'  =  </amrbnr',  by  §§  170.  115, 

=  <fa^  .  </b^'  =  as ' m  •  b1  ' ",  by  §§  120,  176. 

(2)  (amb»)~'=  — J— =    r     l    ^     =a"'"-ft"'"",  by  §  177  and  (1). 

Hence  (ambn)p  =  apmbpn  for  all  rational  values  of  m,  n,  and  p. 

(n»*\  p        amP 
— )    =  —  for  all  rational  values  of 

m,  n,  ana  /),  since 

(^)"  =  (a™  ■  b  n)P  =  amP  ■  b-'P  =  — ,  §  177  and  Law  IV. 

V  (>"  I  bnP 

180.  From  Laws  III,  IV,  and  V,  it  follows  that  any  mono- 
mial is  affected  with  any  exponent  by  multiplying  the  exponent 
of  each  factor'of  the  monomial  by  the  given  exponent. 

Ex.  1.    (ah-2c*)~$  =  a~*  '  V* '  ~2c~%  "  3  =  a~h%~2. 

Ex  2  ( 3  "***) ""  = 3~* a~lx~*  =  h^2 

^   l>!/i  '  b~h~2        $ax* 


Ex.3. 


/  8x9 
\  27  »< 


27  //'\  3  _  27 3  y2  _  3  //'2 


ST9  /  1  •)  r3 


EXERCISES 


Perform  the  operations  indicated  by  the  exponents  in  each 
of  the  following,  writing  the  results  without  negative  expo- 
nents and  in  as  simple  form  as  possible : 


1. 

(M)4- 

5. 

(afV)  *. 

9. 

(¥)~f- 

13. 

(.0009)1 

2. 

(!i)4- 

6. 

25* 

10. 

GV)f- 

14. 

(.027)*. 

3. 

(«)*■ 

7. 

25~l 

11. 

(0.25)*. 

15. 

(32  a-5610)*. 

4. 

(27  a-fl)i 

8. 

25°. 

12. 

(0.25)"*. 

16. 

8*  •  4_i 

146  EXPONENTS  AND   RADICALS 

17.  («\     .  19.    (J2)   *(A)    '•  (        *V     'Ul)     • 

18.  (27aVV^-J.     20.   tytf.(ttt)-*.  WvJUry 
23.   \/81a-468(-27a3&-c)-*.       26-   Vl6a-*&'-".  -\/8a3&-6. 

24  w-^y*.    27  (_2-5„-,,-T5(_o-i(r^-.), 

\m-7J  v  V"     / 

/a»&-"\*  .  Art-\-\  28.  /81r-^Y9^-^V^ 

25-  (^J    A^J  \,626r-A;  i,2or^V 

29.  Prove  Law  III  in  detail  for  the  following  cases: 

(1)  (a»)~«,    (2)  (a~»)«,    (3)  (a"-)""*- 

30.  Prove  Law  IV  in  detail  for  the  following  cases : 

(1)  (ar*b-f)~;    (2)  (a~kb-lf\ 
Multiply : 

31.  ar2  +  x-^y-1  +  y~2  by  ar1  — jr1. 

2  II  S  1  1 

32.  a-3  —  x*y*  + y3  by  x3  +  y* 

4  8      1  22  13  4  1  1 

33.  as*  +  **y*  +  «ay*  +  •*■■'' //■'■  +^/5  by  Xs  — y=. 

34.  \  (/;  +  a/62  by  ^e?—  VW. 

35.  as  ■'  +  x%y%  +  y*  by  a?»  —  yf- 

36.  as  —  Sajty^+SasV-1  —  if*  l,y  a!*  —  2a?*y~*  +  y~1- 

37.  a>*  +  ■'•.'/'  +  *V  +  y*  by  •*"-  -  y* 

Divide  : 

n.  8  11  14  i,  i  i 

38.  /-  —  .(•' •  n  +  ./■-//  —  as%1  +  «:!Z/3  —  ?/>  by  x'—x3i/  +  y. 

39.  3 a£  _  a&f  +  4a62- 3 ah  +  &*  - 4 6s  by  3  a*  -  b*  +  4 &2. 

40.  .c-  -  3  .'■■'  +  G  as*  —  7  x  +  T>  as*  —  .°»  x1*  +  1  by  .r'  —  a^  +  1 . 

41.  ixh-2-17xh2  +  16x~h%by  2  xl- -  b'2  -  i  x~h\ 


REDUCTION   OF  RADICAL   EXPRESSIONS  147 

Find  the  square  root  of : 

42.  4  x2  —  4  xy%  +  4  xz~*  +  y%  —  2  yh~%  +  z~\ 

43.  «"*  —  2  aM  +  Z>3  +  2  a'^c2  4-  c4  —  2  bh-. 

44.  6" *  -  2  &~M  +  c*  +  2  ZT^d*  +  2  &~le~*  _  2  <&#  +  d* 

+  2  dV^  —  2  cV*  +  e_1. 
Find  the  cube  root  of  :  45    i  as  _  |  a*hl  +  Qab_$  &f . 


1,  * 


46.  ac-3a5  +  5a3-3a-l.   47.  a"1  4-3a_t&5  +  3 a"T6T  +  65. 

REDUCTION   OF   RADICAL   EXPRESSIONS 

181.  An  expression  containing  a  root  indicated  by  the  radical 
sign  or  by  a  fractional  exponent  is  called  a  radical  expression. 
The  expression  whose  root  is  indicated  is  the  radicand. 

3/—  2  . 

E.g.  v5  and  (1  -f  x)3  are  radical  expressions.  In  each  case  the 
index  of  the  radical  is  3. 

The  reduction  of  a  radical  expression  consists  in  changing 
its  form  'without  changing  its  value. 

Each  reduction  is  based  upon  one  or  more  of  the  Laws  I  to 
V,  §  175,  as  extended  in  §  179. 

182.  To  remove  a  factor  from  the  radicand.  This  reduction  is 
possible  only  when  the  radicand  contains  a  factor  which  is  a 
perfect  power  of  the  degree  indicated  by  the  index  of  the  root, 
as  shown  in  the  following  examples : 

Ex.1.    V72  =  V3(TT2  =  VM  V2  =  6  V2. 

Ex.  2.   (aWyrf  =  ("Y  •  z2)*  =  («V)'  ■  (**)*  =  ay*x%. 

This  reduction  involves  Law  IV,  and  may  be   written  in 

symbols  thus  :  r 

v  xhy  =  v  xkr  V/  =  *  V/- 


148  EXPOXEXTS  AND   RADICALS 

EXERCISES 

In  the  expressions  on  p.  154,  remove  factors  from  the  radi- 
cands  where  possible. 

In  the  case  of  negative  fractional  exponents,  first  reduce  to  equiva- 
lent expressions  containing  only  positive  exponents. 

183.  To  introduce  a  factor  into  the  radicand.  This  process 
simply  retraces  the  steps  of  the  foregoing  reduction,  and  hence 
also  involves  Law  IV. 

Ex.  1.    6  V2  =  VW2  ■  V2  =  VW^2  =  V72.  See  §  112 


Ex.  2.    ay'2xJ  =  V(a//'2)3-  Vx'2  =  V(ay'2)3x-  =  Va3yGx'2. 
Ex.  3.   x  Vy  =  y/xr  Vy  —  Vx*y. 

EXERCISES 

In  the  expressions  on  p.  154,  introduce  into  the  radicand  any 
factor  which  appears  as  a  coefficient  of  a  radical. 

184.  To  reduce  a  fractional  radicand  to  the  integral  form. 
This  reduction  involves  Law  IV  or  Law  V,  and  may  always  be 
accomplished. 

Ex.  1.    v'|  =    vT|  =  VZV.15=  |\/15.  Law  IV 

Ex.2.    (<l^)K(^^)l=i«^n\=(«*-^1.   LawV 
\a  +  hi  \(u  +  l>y2J         [(a  +  hyji  a+b 

Ex.  3.     — -  =  Ai1—  =  ^/y  _  o. 
</5        *  J 


T  t    ,  ,  r\a       r  abrl      Vabr-1      1  ,.— 

In  svmbols,  we  have  \/    =  \j— -L —  =  — — —  =  7  -sjabr~  K 
*  0        *     or  vA^         0 


EXERCISES 


In  the  expressions  on  p.  154,  reduce  each  fractional  radicand 
to  the  integral  form. 

In  case  negative  exponents  arc  involved,  firsl  reduce  to  equivalent 
expressions  containing  only  positive  exponents. 


REDUCTION   OF  RADICALS  149 

185.  To  reduce  a  radical  to  an  equivalent  radical  of  lower  index. 
This  reduction  is  effective  when  the  radicand  is  a  perfect  power 
corresponding  to  some  factor  of  the  index. 

Ex.1.    ^/8  =  8''=8(^)  =  (8^  =  2^  =  V2. 


Ex.  2.    \/a2  +  2  ab  +  b2  =  yj Va2  +  2  ab  +  62=  Va  +  6. 

This  reduction  involves  Law  III  as  follows : 
ii  n        ^ 

from  which  we  have 

rVx  =  </i'x  =  i/  <jx.  See  §  1 14 

By  this  reduction  a  root  whose  index  is  a  composite  number 
is  made  to  depend  upon  roots  of  lower  degree. 

E.g.  A  fourth  root  may  be  found  by  taking  the  square  root  twice; 
a  sixth  root,  by  taking  a  square  root  and  then  a  cube  root,  etc.  In 
the  case  of  literal  expressions  this  can  be  done  only  when  the  radicand 
is  a  perfect  power  of  the  degree  indicated  by  the  index  of  the  root. 

But  when  the  radicand  is  expressed  in  Arabic  figures,  such  roots  may 
in  any  case  be  approximated  as  in  §  127. 

EXERCISES 

In  the  expressions  on  p.  154,  make  the  reduction  above  indi- 
cated where  possible. 

In  the  case  of  arithmetic  radicands,  approximate  to  two  places  of  deci- 
mals such  roots  as  can  be  made  to  depend  upon  square  and  cube  roots. 

186.    To  reduce  a  radical  to  an  equivalent  radical  of  higher  index. 

This  reduction  is  possible  whenever  the  required  index  is  a 
multiple  of  the  given  index.     It  is  based  on  Law  III  as  follows  : 

r  r    t  rt 

x~s  =  (xs)t  =  xst.  See  §  179 

/-  1  1    I  R  6, 

Ex.1.    Va .  =«2  =  (a2)3  =a6  =  Va3. 
Ex.  2.    Vb  =  b*  =  b%  =  ^6*. 


150  EXPONENTS  AND   RADICALS 

Definition.  Two  radical  expressions  are  said  to  be  of  the 
same  order  when  their  indicated  roots  have  the  same  index. 

By  the  above  reduction  two  radicals  of  different  orders  may 
be  changed  to  equivalent  radicals  of  the  same  order,  namely,  a 
common  multiple  of  the  given  indices. 

E.g.     Va  and  Vb  in  Exs.  1  and  2  above. 

EXERCISES 

In  Exs.  3,  4,  6,  17,  18,  23,  28,  30,  on  p.  154,  reduce  the 
corresponding  expressions  in  the  first  and  second  columns  to 
equivalent  radicals  of  the  same  order. 

187.  In  general,  radical  expressions  should  be  at  once 
reduced  so  that  the  order  is  as  low  as  possible  and  the  radicand 
is  integral  and  as  small  as  possible.  A  radical  is  then  said  to 
be  in  its  simplest  form. 

ADDITION    AND   SUBTRACTION   OF   RADICALS 

188.  Definition.  Two  or  more  radical  expressions  are  said 
to  be  similar  when  they  are  of  the  same  order  and  have  the 
same  radicands. 

E.q.  3V7  and  5V7    are   similar   radicals   as   are   also   aVx4   and 

i,A.  ' 

If  two  radicals  can  be  reduced  to  similar  radicals,  they  may 
be  added  or  subtracted  according  to  §  10. 

Ex.  1.     Find  the  sum  of  \8,  \  50,  and  V98. 

By  §  182,  V8  =  2V2,  V50  =  5V2,  and  V98  =  7V2. 

Hence  Vs  +  V50  +  V9S  =  2  v/2  +  5  V2  +  7V2  =  14 V2. 

Ex.  2.    Simplify  V|  -  V20  +  V3~I. 

By  §  1 8 1.  \  !       1 V5,  V20  =  2  a  o,  V3i  =  >/v  =  4  v  '  =  ••  %  •"'• 

Hence  V\  -  V20  +  \'%  =  l  V5  -  2  Vo  +  |  Vo  =  -  V3. 


MULTIPLICATION   OF  RADICALS  151 

If  two  radicals  cannot  be  reduced  to  equivalent  similar  radi- 
cals, their  sum  can  only  be  indicated. 

E.g.     The  sum  of  V'J  and  vo  is  V2  +  Vo. 

Observe,  however,  that 

VIo  +  Vij  =  V2  •  v?)  +  ^  ■  V3  =  V2  ( V5  +  \/3). 

EXERCISES 

(a)  In  Exs.  1,  2,  5,  7,  8,  19,  20,  21,  p.  154,  reduce  each 
pair  so  as  to  involve  similar  radicals  and  add  them. 

(b)  Perform  the  following  indicated  operations : 

1.  V28  +  3V7-2V63.  5.    V|+ V63  +  5V7. 

2.  a/24  -  ^81  -  -v^.  6.    V99-11VS+ V44. 

3.  Va?  +  vo"  -  \/32a.  7.    2  VJ  +  3  Vf  +  Vl75. 

4.  2V48-3VI2  +  3VJ.  8.    V=b  +  G  V}  -  Vl2. 


9.    V9  +  V27  +  V324. 


10.    ;«2  +  Va3  +  a2b  —  V(a2  —  b'-)(a  —  b). 

MULTIPLICATION    OF    RADICALS 

189.  Radicals  of  the  same  order  are  multiplied  according  to 
§  120  by  multiplying  the  radicands.  If  they  are  not  of  the  same 
order,  they  may  be  reduced  to  the  same  order  according  to  §  186. 

E.g.     y/a-y/b  ■=  orfe*  =  asb^  =  Va3  •  Vb2  =  yVasb'2. 
In  many  cases  this  reduction  is  not  desirable.     Thus,  Vx*-  Vy3  \s 
written  x3y^  rather  than  vx*y9. 

Radicals  are  multiplied  by  adding  exponents  when  they  are 
reduced  to  the  same  base  with  fractional  exponents,  §  176. 

E.g.     a/72  •  Vx3  =  x%  .  x%  =  x%+ *  =  x&. 


152  EXPONENTS   AND   RADICALS 

190.    The  principles  just  enumerated  are  used  in  connection 
with  §  10  in  multiplying  polynomials  containing  radicals. 

Ex.  1.   3  V2  +  2  V5  Ex.  2.   3  V2  +  2  V5 

2  V2  -  3  V5  3  V2  -  2  Vo 


6-2    +  4  VlO  9-2    +6V10 

-  9  VlO  -6-5  -  6  VlO  -4-5 

12      -  5  VlO  -  30  18  -  20 

Hence  (2  V2  +  2  V5)(2  V2  -  3  V5)  =  -  18  -  5  VlO, 

and  (3  V2  +  2  V5)(3  V2  -  2  V5)  =  18  -  20  =  -2. 

EXERCISES 

(a)  In   Exs.   21  to   38,  p.   154,  find   the  products    of   the 
corresponding  expressions  in  the  two  columns. 

(b)  Find  the  following  products: 

i.  (3  +  Vii)(3  -  Vii). 

2.  (3  V2  +  4  V5)(4  V2  -  5  V5). 

3.  (2  +  V3  +  \/5)(3  +  VB  -  V5). 

4.  (3  V2  -  2  V18  +  2  V7)(2  V2  -  Vl8  -  V7). 

5.  (Va  -  V6)(Va  +  V6)(a2  +  ab  +  b2). 

6.  (Vvl3+3)(VVl3-3). 

7.  (V2  +  3  VB)(V2  +  3  V6). 

8.  (3  a  -  2  V.;i-)(4  a  +  3  Vx). 

9.  (3  V3  +  2  V6  -  4  V8)(3  V3  -  2  V6  +  4  V8). 
(Va  +  V6  -  ^Jc)(y/a  -  V6  +  Vc). 


10. 


DIVISION   OF  RADICALS  153 

11.  (a  —  Vb  —  -Vc)(a  +  V6  +  Vc). 

12.  (2Vl  +  3V!  +  4V!)(2V!-5Vf). 

13.  (Va2  +  ^(Va2  +  \/a^  +  ^3). 

14.  (V^3-i/3)3. 

DIVISION  OF  RADICALS 

191.  Radicals  are  divided  in  accordance  with  Laws  II  and  V. 
That  is,  the  exponents  are  subtracted  when  the  bases  are  the 
same,  and  the  bases  are  divided  when  the  exponents  are  the  same. 
See  §§  179,  121. 

-n  -.  5/—,  IS  2  3  2.3  .II 

Ex.  1.    yV  h-  V.s3  =  a3  -5-  a;2  =  a;5  2  =  x  x °. 


Ex.  2.   a?*  -s-  y*  =  ( -)*=  {xy~ly=  <fx2y~2. 


Ex.3.    Va-V6  =  a«-6S  =  (~f  =  Va36-2. 


EXERCISES 

(a)    In  each  of  the  Exs.  1   to  20,  on   p.  154,   divide  the 
expression  in  the  first  column  by  that  in  the  second. 

(6)    Perforin  the  following  divisions  : 

1.  ( Vu3  +  2  Vo*  -  3  Va)  h-  6  Va. 

2.  (Va  +  V&  -  c)  -*-  Vc. 

3.  (2  V9  +  3  Vl2  -  4  Vl5)  -h  V3. 

4.  (4  V7  -  8  V2i  +  6  V42)  -  2  V7. 


154  EXPONENTS  AND  RADICALS 

EXERCISES 

1.  3V45,  2V125.             21.  3(50)2,  4(72)1 

2.  a*,  ak                      22.  a*,  ai 

3.  x-5,  x?.                        23.  <'.<S  fcci 

4.  3Vx-2?/,  2i/x\    ■             24.  a*&^  aV. 

>r           7    3/     o,s  3/— 5^9                             „                  i     4,7                            3      2,1 

5.  dva-b-,  cvao-              25.  Dr'tt-h,  m*n*ls. 


6.  7V(a  +  Z>)3,    ll\/(a-6)6.     26.   5VaW,      3Va3//cl;f. 

7.  A^a4,  al                       27.    -Va8  </&">,  V  69  a/o". 

/ — s  3 

8.  nv»l,  Wl5. 

9.  Vf,  Vf 

10    Vi?  V^ 

1U.       V    72,  v  54- 

11.  Vf|,  Vff. 

12.  — ,  rVF. 
6"2 

13.  a-2&8, 


a° 


28. 

8*, 

165 

29. 

25~\ 

125-3. 

30. 

9t, 

8i 

31. 

(*)*, 

(A)"*- 

32. 

5  a*3,?/5 

x3y5z- 

„       .5     8  5,3 

3a  -/--         i  a  *&  5 

00.       , 

1,  _i 


64  4a"*&~*       5a~^6"s 

14.    A  '£!£!.  34    dd-*  A0 


»H-7'                                4-                                        '        ,-i    -i'  1 

3ft-|  "  dVr-  3 

15-   T7^'  ^~*  35-   5(a+&)"3'  3(a +  &)-*. 

16.    ^  ^J.  36-    ("64>*  ~64i- 

IT.    tt  WL  3?-(^  <*>* 


-  3  r-3 

is.  s/^,  -  &\  ; 

19.    IS1.  \ .'32.  39.    -v/4tt8-12a26+12a62-468. 


1         V»3 


SO.    Vl2,  48*.  40.  V(3a-2&)(9a2-4&2). 


BATIONALIZING    THE  DIVISOR  155 

192.  Rationalizing  the  divisor.  In  case  division  by  a  radical 
expression  cannot  be  carried  out  as  in  the  foregoing  examples, 
it  is  desirable  to  rationalize  the  denominator  when  possible. 


Ex.  1. 


Ex. 


V2^  V2-  Vo  =  VlO 

^5~V5'V5        5 

Va  -\/a(y/a  +  Vo)         _  a  +  Vab 

Va-V&      ( Va -  V6)( Va  +  V&)         a-6 


Evidently  this  is  always  possible  when  the  divisor  is  a  mono- 
mial or  binomial  radical  expression  of  the  second  order. 

The  number  by  which  numerator  and  denominator  are  multi- 
plied is  called  the  rationalizing  factor. 

For  a  monomial  divisor,  Vx,  it  is  Vx  itself.  For  a  binomial  divisor, 
Va;±  Vy,  it  is  the  same  binomial  with  the  opposite  sign,  y/xT  Vy. 

EXERCISES 

Reduce  each  of  the  following  to  equivalent  fractions  having 
a  rational  denominator. 


1. 


2  _ 

V5' 

7 

V5 

+  V3 

V27 

V3 

+  VH 

2  — 

V7 

2  + 

V7 

V2 

-V3 

Va;  +  Vy_ 
Va;  —  Vy 

3V3-2V2 
3  V3  +  2  V2 


3. 


4.    = ^=-  9. 


Va2  +  1  —  Va 

2-l 

Va2  —  1  +  V« 

Vx  +  1  +  Va;  - 

+  1 
-1 

V#  +  1  —  Va;  ■ 

-1 

Va  —  b  —  Va 

+-■& 

10. 

V2  +  V3  Va  -  &  +  Va  +  b 


156  EXPONENTS  AND   RADICALS 

193.    In    finding     the    value    of     such    an     expression     as 
— —^ — =,  the  approximation  of  two  square  roots  and  division 

by  a  decimal  fraction  would  be  required.     But  v  i  "*"  v^  equals 

114-2V21  V7-V3 

— C— — —  which  requires  only  one  root  and  division  by  the 

integer  4. 

EXERCISES 

Find  the  approximate  values  of  the  following  expressions  to 
three  places  of  decimals. 

1     3V5  +  4V3  g     7V5  +  3V8 

V5  -  V3 

2.  ^ 


V7  -  V2 

4V3 

V3  -  V2 


2Vo  - 

-3V2 

5V19 

-3V7 

3V7 

-  \  19 

3V2- 

-V5 

V5- 

6  V2 

5V^- 

-7  a/13 

4     11V5-3V3  g 

2  V5  +  V3  3  v  13  -  7  V6 

194.  Square  root  of  a  radical  expression.  A  radical  expression 
of  the  second  order  is  sometimes  a  perfect  square,  and  its 
square  root  may  be  written  by  inspection. 

E.g.  The  square  of  Va  ±  Vb  is  a  +b  ±  2  Vab.  Hence  if  a  radical 
expression  can  be  put  into  the  form  x  ±  2v/.  where  x  is  the  sum  of 
two  numbers  a  and  b  whose  product  is  y,  then  va  ±  Vb  is  the  square 
root  of  x  +  2  v^. 

Example.     Find  the  square  root  of  5  -f-  v24. 

Since  o  +  V24  =  5  +  2V6,  in  which  5  is  the  sum  of  2  and  3,  and  6  is 
their  product,  we  have  V5  +  y/94,  =  V2  +  VS. 


IMA  GIN  A  HIES  157 

EXERCISES 

Find  the  square  root  of  each  of  the  following : 

1.  3  -  2  V2.  3.   8  -  V60.  5.  24  -  6  Vf. 

2.7  +  V40.  4.    7  +  4  V3.  6.    28  +  3  Vl2. 

195.    Radical  expressions  involving  imaginaries.     According  to 
the   definition,  §  112,    (V^T)2  =  -  1.        Hence,     (V  —  l)3 

=  (V^T)2V^T  =  -V^T    and    (V^I)4=(V^1)2(V^1)2 
=  (-l)(-l)=  +  l. 

The  following  examples  illustrate  operations   with   radical 
expressions  containing  imaginaries. 

Ex.1.    V"zr4  +  V"=r9  =  V4V:rT  +  V9  V^l 

=  (2  +  3)  V^l  =  5  V^l. 

Ex.  2.    V^4  •  V^9  =  v'4  •  V9  •  ( V^)2  =  -  2  •  3  =  -  6. 

Ex  3     y/~~1  =  ^  ^^  =  V^  =  2 
V^~9      V9V^1      V9      3 

Ex.  4.    V^2  •  V^3  •  V^6  =  V2  •  V3  •  V6  ■  (V^T)3 

=  -V36V^T==-6V^T. 
Ex.  5.    Simplify  ($  +  | V^3)3. 

We    are   to   use    %(l+\/~3\/—  1)   three   times   as   a  factor.      Re- 
serving (J)3  =  £  as  the  final  coefficient,  we  have, 

1  +  V3  V^l  -  2  +  2V3  V^T 

i  +  va  V3T  i  +   V3V^1 

1  +  V3  V^T  -  2  +  2  V3  i/-T 

V3~V^T-3  -2V3V3l_6 


1  +  2  V3  V  -  1  -  3  -  2  -  6  = 

Hence   Q  +  £  V^3)8  =  i(-  8)  =  -  1. 


158 


EXPONENTS  AND   RADICALS 


EXERCISES 

Perform  the  following  indicated  operations. 


1.  V-  16  +  V^  +  V-25. 

2.  V^4 


■V—  x2. 

3.  3  +  5V:rT-2y 

4.  (2  +  3V^l)(3  +  2V^l). 

5.  (2  +  3V^I)(2-3V^T). 

6.  (4.4-5V:r3)(4-5V:r3). 

7.  (2  V2 -  3  y/r^S) (3  V3  +  2  V=2). 

8.  (V^3  +  V^2)(V:::3-V^2). 

9.  (3  Vo  +  2  V^7)(2  V5  -  3  V^T). 
10.    (-|_+V^3)(-l-iV^3)'-'. 

Rationalize  the  denominators  of 


11. 


12. 


13. 


1-V-l 

3 

V3+V^l 

1-V-l 

l+V-i 


14. 


15. 


16. 


V2+V" 

-3 

V2-V" 

^3 

5 

2  -  3  V  - 

5 

x  4-  y  V^ 

^1 

xV— 1  —  # 


17.  Solve  ic4  —  1  =  0  by  factoring.    Find  four  roots  and  verify 
each. 

18.  Solve  ar'  +  1  =  0  by  factoring  and  the  quadratic  formula. 
Find  three  roots  and  verify  each. 

19.  Solve  a?  —  1  =  0  as  in  the  preceding  and  verify  each  root. 

20.  Solve  x6  —  1  =  0  by  factoring  and  the  quadratic  formula. 


EQUATIONS   CONTAINING   RADICALS  159 

SOLUTION    OF   EQUATIONS    CONTAINING   RADICALS 

196.  Many  equations  containing  radicals  are  reducible  to 
equivalent  rational  equations  of  the  first  or  second  degree. 

The  method  of  solving  such  equations  is  shown  in  the  fol- 
owing  examples. 


Ex.  1.    Solve  1  +  Va-  =  V3  +  x.  (1) 

Squaring  and  transposing,        2vs  =  2.  (2) 

Dividing  by  2  and  squaring,  x  =  1.  (3) 

Substituting  in  (1),  1  +  1  =  V3  +  1  =  2. 

Observe  that  only  principal  roots  are  used  in  this  example. 

If  (1)  is  written  1  +  Var  =  -  V3  +  x,  (4) 

then  (2)  and  (3)  follow  as  before,  but  x  =  1  does  not  satisfy  (4).  In- 
deed algebra  furnishes  no  means  whereby  to  obtain  a  number  which  will 
satisfy  (4). 


Ex.  2.    Solve  Vx  +  5  =  *  - 1.  (1) 

Squaring  and  transposing.         x'2  —  3  x  —  4  =  0.  (2) 

Solving,  x  —  4  and  x  —  —  1. 


x  =  4  satisfies  (1)  if  the  principal  root  in  Vx  +  5  is  taken.  x=  —I 
does  not  satisfy  (1)  as  it  stands  but  would  if  the  negative  root  were 
taken. 


tt      q     «  i  V4.r  +  1-  V3a;-2_1  m 

Ex.3.   Solve     — ===== ==•  (.-1; 

V4a  +  l  +  V3a;-2      5 

Clearing  of  fractions  and  combining  similar  radicals. 


2V4x-+  1  =  3V3x  -2.  (2) 

Squaring  and  solving,  we  find  x  =  2. 

This  value  of  x  satisfies  (1)  when  all  the  roots  are  taken  positive  and 
also  when  all  are  taken  negative,  but  otherwise  not. 


EXPOXENTS  AXD   RADICALS 

Zx~l    -1.  (1) 

The  fraction  in  the  second  member  should  be  reduced  as  follows : 

3x-l        (VWx  -  l)(VS~x  +  1)         r— 

— = —  — =  V3x  +  l. 

V3  x  -  1  V3  x  -  1 


Hence,  (1)  reduces  to     V2  x  +  3  =  V3x  +  1  -  1  =  a/3x.  (2) 

Solving,  x  =  3,  which  satisfies  (1). 

If  we  clear  (1)  of  fractions  in  the  ordinary  manner,  we  have 


(V3  x  -  1)  V2  x  +  3  =  -  V3  x  +  3  x.  (2') 

Squaring   both    sides   and  transposing   all  rational    terms  to  the 

second  member,  

2  x  y/3  x-6  Vfilc  =  3  x2  -  8  x  -  3.  (3) 

Factoring  each  member, 

2(x  -  3)  V¥x  =  (x  -  3) (3  x  +  1),  (4) 

which  is  satisfied  by  x  =  3. 

Dividing  each  member  by  x  —  3,  squaring  and  transposing,  we  have 
9ar2-6a;+l=  (3  x  -  1)-  =  0,  (5) 

which  is  satisfied  by  x  =  3. 

Equation  (1)  is  not  satisfied  by  x  =  i,  since  the  fraction  in  the 
second  member  is  reduced  to  {J  by  this  substitution.  Sec  i  ."in. 
The  root  x  =  ±  is  introduced  by  clearing  of  fractions  without 
first  reducing  the  fraction  to  its  lowest  terms,  Vox— 1  being  a 
factor  of  both  the  numerator  and  the  denominator.     See  §  165. 

Ex.5.     Solve     (]~x    -V^=    *~3   •  (1) 

V0  —  x  V&  —  3 

Reducing  the  fractions  by  removing  common  factors,  we  have 

V(fZ  x  -  y/3  =  vT^~3.  (2) 

Squaring,  transposing,  and  squaring  again, 

x2-9x+18  =  0,  (3) 

whence  x  =  3,    x  =  6. 

But  neither  of  these  is  a  root  of  (1).      In  this  ease  (1)  has  no  root. 


EQUATIONS    CONTAINING    RADICALS  161 

197.  In  solving  an  equation  containing  radicals,  we  note  the 
following : 

(1)  If  a  fraction   of   the  form       ®~         is   involved,  this 

V«  —  V6 
should  be  reduced  by  dividing  numerator  and  denominator  by 
s/a  —  V&  before  clearing  of  fractions. 

(2)  After  clearing  of  fractions,  transpose  terms  so  as  to 
leave  one  radical  alone  in  one  member. 

(3)  Square  both  members,  and  if  the  resulting  equation  still 
contains  radicals,  transpose  and  square  as  before. 

(4)  In  every  case  verify  all  results  by  substituting  in  the 
given  equation.  In  case  any  value  does  not  satisfy  the  given 
equation,  determine  whether  the  roots  could  be  so  taken  that 
it  would.     See  Ex.  3. 

EXERCISES 

Solve  the  following  equations  : 


1.    Va-2  +  7a;-2-Va;2-3a,'  +  6  =  2. 


2.    V3?/-V3//-7 


b 


V3.V-7 


fr/-l    =  Vfy-l    .    t  8.    V5^  +  l  =  l_    5^~-1 

V&v  +  l  2  V5a;-fl 


4.    V5aj-19+V3a;  +  4  =  9.      „     4 

9-  =V3. 


Va;2  +  a2  —  x 
-Vx-  +  a2  +  x 


x 


4  +  g  +  V8a;  +  ar=1> 


Va  +  Vox + x'1  =  Va —  Va\  4  +  x  —  V8«  +  x2 


7. 


y-*   =^y-^~l+2Vi.  ii.    "-*  4-  -T+ft-=v^ 

V?/+V?  °  Va  — a-     Va-+a 

12.       x~a    _  =  V*  +  V"  +  2V«., 

V^  — Va  -1  ■ 


162  EXPONENTS   AND   RADICALS 

13.    —       a    —  V  m  —  n  = • 

Vm  —  y  V#  —  n 

i4.  2v;n7i  +  3y^  =  7f'±;K 

Va-  —  a 


1 5 .    V2  g  +  7  +  V2.r  +  14  =  "V4  x  +  35  +  2  V4  x2  +  42  a;  -  21. 


16.  Vo;  -3  +  Vw  +  9"=  Va;  + 18  +  Va; -  6. 

17.  V.7+7  -  Va;  — 1  =  Vx  +  2  +  V.c  -  2. 


18.  a V#  +  b  —  c  V&  —  y  =  V6  (a2  +  c2) . 

1 9.  i/  V//  —  c  —  Vy"  +  c3  +  c V#  +•  c  =  0. 

20.  v?+y*»^vw  |  v^ 

Vm      Va;       Vw       Vm 

12 


21.   \14  +  Va?+\6-v*= 


\  6  -  \  x 


91 
22.     V  3  g  +  V3  a?  +  13  =  —  =. 

V3  *  + 13 


23.    V6g  +  3  +  Vg  +  3  =  2g+3. 


24.    v  x  -  a  +  Vft  -  x  =  Vft 


0  -     Va;  —  a+ Va;— ft         /a;— a 

^5.     —  -  —         =="V ■ 

\ x      a       \ x—b       *x—b 


26.    V(g  -  l)(g  -  2)  +  V(g— 3)(x  -  4)  =  V2. 


27.    V2  x  +  2  +  V7  +  6 .«  =  V7a;+72. 


2 8 .    V 2  ( i  bx  +  Va2  —  6a;  =  Va2  +  bx. 
a  -+-g-f  \  a*      .'•-'      c 


29 


+  .'•  —  Va2  • 


30.    Va?  -  2  .r  +  4  +  V3  ar  +  6  a;  +  12  =  2  Va;2  +  g  +  10. 


PROBLEMS 


163 


PROBLEMS 

1.  Find  the  altitude  drawn  to  the  longest  side  of  the  tri- 
angle whose  sides  are  6,  7,  8. 

Hint.  See  figure,  p.  235,  E.  C.  Calling  %  and  8  —  x  the  segments 
of  the  base  and  h  the  altitude,  set  up  and  solve  two  equations  involv- 
ing x  and  it. 

2.  Find  the  area  of  a  triangle  whose  sides  are  15,  17,  20. 
First  find  one  altitude  as  in  problem  1 . 

3.  Find  the  area  of  a  triangle  whose  base  is  16  and  whose 
sides  are  10  and  14. 

4.  Find  the  altitude  on  a 
side  a  of  a  triangle  two  of 
whose  sides  are  a  and  a  third  b. 

A  three-sided  pyramid  all  of 
whose  edges  are  equal  is  called  a 
regular  tetrahedron.  In  Figure  10 
AB,  A  C,  AD,  BC,  BD,  CD  are  all 
equal. 

5.  Find  the  altitude  of  a 
regular  tetrahedron  whose  edges 
are  each  6.  Also  the  area  of  the 
base. 

Hint.  First  find  the  altitudes  AE  and  DE  and  then  find  the  alti- 
tude of  the  triangle  AED  on  the  side  DE,  i.e.  find  AF. 

6.  Find  the  volume  of  a  regular  tetrahedron  whose  edges 
are  each  10. 

The  volume  of  a  tetrahedron  is  \  the  product  of  the  base  and 
the  altitude. 

7.  Find  the  volume  of  a  regular  tetrahedron  whose  edges 
are  a. 

8.  In  Figure  10  find  EG  if  the  edges  are  a. 

9.  If  in  Figure  10  EG  is  12,  compute  the.  volume. 

Use  problem  S  to  find  the  edge,  then  use  problem  7  to  find  the 
volume. 


Fig.  10. 


164  EXPONENTS  AND  RADICALS 

10.  Express  the  volume  of  the  tetrahedron  in  terms  of  EG. 
That  is  if  EG  =  b,  find  a  general  expression  for  the  volume  in 
terms  of  b. 

11.  If  the  altitude  of  a  regular  tetrahedron  is  10,  compute 
the  edge  accurately  to  two  places  of  decimals. 

12.  Express  the  edge  of  a  regular  tetrahedron  in  terms  of 
its  altitude. 

13.  Express  the  volume  of  a  regular  tetrahedron  in  terms 

of  its  altitude. 

14.  Express  the  edge  of  a  regular  tetrahedron  in  terms  of 

its  volume. 

15.  Express  the  altitude  of  a  regular  tetrahedron  in  terms 
of  its  volume. 

16.  Express  EG  of  Figure  10  in  terms  of  the  volume  of  the 
tetrahedron. 

17.  Find  the  edge  of  a  regular  tetrahedron  such  that  its  vol- 
ume multiplied  by  V2,  plus  its  entire  surface  multiplied  by  V3, 
is  114. 

The  resulting  equation  is  of  the  third  degree.     Solve  by  factoring. 

18.  An  electric  light  of  32  candle  power  is  25  feet  from  a 
lamp  of  G  candle  power.  "Where  should  a  card  be  placed 
1  iet  ween  them  so  as  to  receive  the  same  amount  of  light  from 
each '.' 

Compare  problem  b5.  p.  141.  Compute  result  accurately  to  two 
places  of  decimals. 

19.  "Where  must  the  card  be  placed  in  problem  18  if  the 
lamp  is  between  the  card  and  the  electric  light? 

Notice  that  the  roots  of  the  equations  in  IS  and  19  are  the  same. 
Explain  what  this  means. 

20.  State  and  solve  a  general  problem  of  which  18  and  19 
are  special  cases. 


PROBLEMS  165 

21.  If  the  distance  between  the  earth  and  the  snn  is  93 
million  miles,  and  if  the  mass  of  the  snn  is  300,000  times  that 
of  the  earth,  find  two  positions  in  which  a  particle  would  be 
equally  attracted  by  the  earth  and  the  sun. 

The  gravitational  attraction  of  one  body  upon  another  varies  in- 
versely as  the  square  of  the  distance  and  directly  as  the  product  of 
the  masses.     Represent  the  mass  of  the  earth  by  unity. 

22.  Find  the  volume  of  a  pyramid  whose  altitude  is  7  and 
whose  base  is  a  regular  hexagon  whose  sides  are  7. 

The  volume  of  a  pyramid  or  a  cone  is  \  the  product  of  its  base  and 
its  altitude. 

23.  If  the  volume  of  the  pyramid  in  problem  22  were  100 
cubic  inches,  what  would  be  its  altitude,  a  side  of  the  base 
and  the  altitude  being  equal  ?  Approximate  the  result  to  two 
places  of  decimals. 

24.  Express  the  altitude  of  the  pyramid  in  problem  22  in 
terms  of  its  volume,  the  altitude  and  the  sides  of  the  base 
being  equal. 

25.  If  in  a  right  prism  the  altitude  is  equal  to  a  side  of  the 
base,  find  the  volume,  the  base  being  an  equilateral  triangle 
whose  sides  are  a. 

The  volume  of  a  right  prism  or  cylinder  equals  the  product  of 
its  base  and  its  altitude. 

26.  Find  the  volume  of  the  prism  in  problem  25  if  its  base 
is  a  regular  hexagon  whose  side  is  a. 

27.  Express  the  side  of  the  base  of  the  prism  in  problem  25 
in  terms  of  its  volume.  State  and  solve  a  particular  problem 
by  means  of  the  formula  thus  obtained. 

28.  Express  the  side  of  the  base  of  the  prism  in  problem  26 
in  terms  of  its  volume.  State  and  solve  a  particular  problem 
by  means  of  the  formula  thus  obtained. 


106 


EXPONENTS  AND   RADICALS 


In  Figures  11  and.  12  the  altitudes  are  each  supposed  to  be 
three  times  the  side  a  of  the  regular  hexagonal  bases. 


Fig.  11. 


Fig.  12. 


29.  Express  the  difference  between  the  volumes  of  the  pyra- 
mid and  the  circumscribed  cone  in  terms  of  a. 

The  volume  of  a  cone  equals  \  the  product  of  its  base  and  altitude. 

30.  Express  a  in  terms  of  the  difference  between  the  volumes 
of  the  cone  and  pyramid.  State  and  solve  a  particular  problem 
by  means  of  the  formula  thus  obtained. 

31.  Express  the  volume  of  the  pyramid  in  terms  of  the  dif- 
ference between  the  areas  of  the  bases  of  the  cone  and  the 
pyramid.  State  a  particular  case  and  solve  by  means  of  the 
formula  first  obtained. 

The  lateral  area  of  a  right  cylinder  or  prism  equals  the  perimeter 
of  the  base  multiplied  by  the  altitude. 

32.  Express  the  difference  of  the  lateral  areas  of  the  cylin- 
der and  the  prism  in  terms  of  a. 

The  following  four  problems  refer  to  Figure  \2.  In  each  case  state 
a  particular  problem  and  solve  by  means  of  the  formula  obtained. 

33.  Express  a  in  terms  of  the  difference  of  the  lateral  areas. 

34.  Express  the  volume  of  the  prism  in  terms  of  the  differ- 
ence of  the  perimeters  of  the  bases. 

35.  Express  the  volume  of  the  cylinder  in  terms  of  the  dif- 
ference of  the  lateral  areas. 

36.  Express  the  sum  of  the  volumes  of  the  prism  and  cylin- 
der in  terms  of  the  difference  of  the  areas  of  the  bases. 


CHAPTER   XI 
LOGARITHMS 

198.  The  operations  of  multiplication,  division,  and  finding 
powers  and  roots  are  greatly  shortened  by  the  use  of  logarithms. 

The  logarithm  of  a  number,  in  the  system  commonly  used, 
is  the  index  of  that  power  of  10  which  equals  the  given  number. 

Tims,  2  is  the  logarithm  of  1U0  since  10-  =  100. 
This  is  written  log  100  =  2. 

Similarly  log  1000  =  3,  since  103  =  1000, 

and  log  10000  =  1,  since  104  =  10000. 

The  logarithm  of  a  number  which  is  not  an  exact  rational 
power  of  10  is  an  irrational  number  and  is  written  approxi- 
mately as  a  decimal  fraction. 

Thus,  log  71  =  1.8092  since  101SG92  =  74  approximately. 
In  higher  algebra  it  is  shown  that  the  laws  for  rational  ex- 
ponents (§  179)  hold  also  for  irrational  exponents. 

199.  The  decimal  part  of  a  logarithm  is  called  the  mantissa, 
and  the  integral  part  the  characteristic. 

Since  10"  =  1,  10l  =  10,  10- =  100,  103  =  1000,  etc.,  it  follows 
that  for  all  numbers  between  1  and  10  the  logarithm  lies  be- 
tween 0  and  1,  that  is,  the  characteristic  is  0.  Likewise  for 
numbers  between  10  and  100  the  characteristic  is  1,  for  num- 
bers between  100  and  1000  it  is  2,  etc. 

200.  Tables  of  logarithms  (see  p.  170)  usually  give  the  man- 
tissas only,  the  characteristics  being  supplied,  in  the  case  of 
whole  numbers,  according  to  §  199,  and  in  the  case  of  decimal 
numbers,  as  shown  in  the  examples  given  under  §  201. 

167 


108  LOGARITHMS 

201.  An  important  property  of  logarithms  is  illustrated  by 
the  following: 

From  the  table  of  logarithms,  p.  170,  we  have  : 

log  376  =  2.5752,  or  376  =  102-5752.  (1) 

Dividing  both  members  of  (1)  by  10  we  have 

37.6  =  102-5752  -  101  =  lO2-5752-1  =  10157*2. 
Hence,  log  37.6  =  1.5752, 

Similarly,  log  3.76  =  1.5752  —  1  =  0.5752, 

log  .376  =  0.5752  —  1,  or  1.5752, 
log  .0376  =  0.5752  -  2.  or  2.5752, 

where  1  and  2  are  written  for  —  1  and  —  2  to  indicate  that  the  char- 
acteristics are  negative  while  the  mantissas  are  positive. 
Multiplying  (1)  by  10  gives 

log  3760    =  2.5752  +  1  =  3.5752, 
and  log  37600  =  2.5752  +  2  =  4.5752. 

Hence,  if  the  decimal  point  of  a  number  is  moved  a  certain 
number  of  places  to  the  right  or  to  the  left,  the  characteristic  of 
the  logarithm  is  increased  or  decreased  by  a  corresponding 
number  of  units,  the  mantissa  remaining  the  same. 

From  the  table  on  pp.  170,  171,  we  may  find  the  mantissas 
of  logarithms  for  all  integral  numbers  from  1  to  1000.  In  this 
table  the  logarithms  are  given  to  four  places  of  decimals, 
which  is  sufficiently  accurate  for  most  practical    purposes. 

E.fj.  for  log  4  the  mantissa  is  the  same  as  that  for  log  40  or  for  log 
400. 

To  find  log .0376  we  find  the  mantissa  corresponding  to  376,  and 
prefix  ilif  characteristic  2.     See  above. 

Ex.  1.    Find  log 876. 

Solution.  Look  down  the  column  headed  N  to  87,  then  along  this 
line  to  the  column  headed  6,  where  we  liud  the  number  9425,  which  is 
the  mantissa.      Hence  log 876  =  2.9425. 


LOGARITHMS  169 

Ex.  2.    Find  log  3747. 

Solution.   As  above  we  find  log  3740  =  3.5729, 
and  log  3750  =  3.5740. 

The  difference  between  these  logarithms  is  11,  which  corresponds  to 
a  difference  of  10  between  the  numbers.  But  3740  and  3747  differ 
by  7.  Hence,  their  logarithms  should  differ  by  -fa  of  11,  i.e.  by  8.1. 
Adding  this  to  the  logarithm  of  3740,  we  have  3.5737,  which  is  the 
required  logarithm. 

The  assumption  here  made,  that  the  logarithm  varies  directly 
as  the  number,  is  not  quite,  but  very  nearly,  accurate,  when  the 
variation  of  the  number  is  confined  to  a  narrow  range,  as  is 
here  the  case. 

Ex.  3.    Find  the  number  whose  logarithm  is  2.3948. 

Solution.  Looking  in  the  table,  we  find  that  the  nearest  lower  loga- 
rithm is  2.3945  which  corresponds  to  the  number  218.     See  §  199. 

The  given  mantissa  is  3  greater  than  that  of  248,  while  the  man- 
tissa of  249  is  17  greater.  Hence  the  number  corresponding  to 
2.3948  must  be  248  plus  T\  or  .176.  Hence,  248.18  is  the  required 
number,  correct  to  2  places  of  decimals. 

Ex.  4.    Find  log  .043. 

Solution.    Find  log  43  and  subtract  3  from  the  characteristic. 

Ex.  5.    Find  the  number  whose  logarithm  is  4.3949. 

Solution.  Find  the  number  whose  logarithm  is  0.3949,  and  move 
the  decimal  point  4  places  to  the  left. 

EXERCISES 

Find  the  logarithms  of  the  following  numbers  : 

1.  491.                 6.   .541.                 11.    .006.  16.  79.31. 

2.  73.5.                 7.    .051.                  12.    .1902.  17.  4.245. 

3.  2485.                8.    8104.                 13.    .0104.  18.  .0006. 

4.  539.7.               9.    70349.               14.    2.176.  19.  3.817. 

5.  53.27.            10.   439.26.             15.   8.094.  20.  .1341. 


170 


LOGARITHMS 


N 

0 

1 

2 

3 

4 

5   6   7 

8 

9 

10 

0000  0U43  0086  0128  0170 

0212  025:;  H204  0334  0374 

11 

0414 

04.:.:; 

0492  0531 

0509 

0607  0645 

0682  0710  0755 

12 

0702 

0828 

0864  0899 

0934 

0969  1004 

1038  1072 

1100, 

13 

1139 

1173 

1206  1239 

1271 

1303  1335 

L367  1300 

1150 

14 

1401 

1492 

152:;  1553  j  1584 

1614  1044 

1673  1703 

1732 

15 

1701 

1790 

1818 

1847  1875 

1903 

1931  1959  L987 

2014 

16 

2H41  2068 

2095 

2122  2148 

2175 

2201  2227 

2353 

2279 

17 

2304  2330 

2355 

2380  2405 

2430 

2455  2480 

2504 

2520 

18 

2553  2577 

2001 

2625  2648 

2672 

2005  27  is 

27  12 

2705 

19 

2788  2810 

2833 

2856  2878 

2900 

2923  20  15 

2967  2989 

20 

3010 

3032 

3054 

3075 

3096 

31  is 

3139  3160 

3181 

520  1 

21 

3222 

324:1 

3263 

3284 

3304 

3324 

5315 

330,5 

33s5 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

354  1 

3560 

3,570 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

370,0 

37S4 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

30  15 

3962 

25 

3979 

3997 

4014  4031 

4048 

lot;  5 

4082 

4099 

1110 

4133 

26 

1 1 51 1 

H66 

4183 

42(10 

4216 

12:12  4210  4205 

42-1 

4298 

27 

4314 

1330 

4346 

1362 

4378 

4393 

4400 

4425 

4440 

4450, 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

1570 

4594 

4609 

29 

4624 

4639 

4654  4669 

4683 

L698 

4713 

172- 

4742 

4757 

.{() 

4771 

4786 

4800  4814 

4829 

1*43 

4857 

4871  4886 

4900 

31 

4914 

1928 

4042 

4055 

4000 

4983 

4907 

5011  5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145  5159 

5172 

33 

5185 

5 19W 

5211 

5224  5237 

5251) 

5203 

5270  5289 

5302 

34 

5315  5328 

5340 

5353  5366 

5378 

5391 

5403 

5410 

5428 

.;.-, 

5441 

5453 

5465  5478  5490 

5502 

5514 

5527 

553,0 

5551 

36 

5563 

5575 

5587  5500  5611 

50,23 

5635 

5047 

5658 

5670 

37 

5682 

5694  5705 

57  17  5729 

5740  5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832  5843 

5855  5866 

5877 

5sss 

5S99 

39 

5911 

5922 

5933 

5944  1  5955 

5966  5!  17  7 

5988 

5000 

6010 

40 

6021 

oo:;i 

6042 

6053 

606  1 

0075 

6085 

0090 

0107 

6117 

41 

6128 

6138 

6149 

6160 

0170 

0180 

619]  0,201 

0212 

0,222 

42 

6232 

6243 

6253 

020.'! 

6274 

02  S( 

0,20  1  030  1 

6314 

03,25 

43 

6335 

6345 

6355 

6365 

6375 

6385 

0305  6405 

0,415 

0125 

44 

6435 

6444 

6454 

6464 

6474 

6484 

0103,  0503 

0,513, 

0,522 

45 

6532 

0.",  4  2 

6551 

6561 

6571 

6580 

c,5:  mi 

6599  6609 

0,0  is 

46 

6628 

6637 

oo4'; 

6656 

6665 

6675 

6684 

6693  0,702 

0712 

47 

0721 

6730 

6739 

0710 

6758 

070,7 

0770 

6785  0,701 

6803 

48 

6812 

6821  6830 

6839  6848 

7857 

osoo, 

6875  6884 

0,so5 

49 

6902  6911  |6920 

6928  j  6937 

6946 

0055 

0004  0072 

oosi 

50 

6990  1  6998 

7(107 

7010  7024 

7033  7042 

7050  7059 

7067 

51 

7076  7084 

7093 

7101  7110 

71  is  7120, 

7135  7143  7152 

52 

7160  7168 

7177 

7185  710:; 

7202  7210 

72  is  7220  723,5 

53 

72  1::  7251 

7259 

7207  7275 

7284  7202 

73.00  7308  7310, 

54 

7324  7332  7340  7348  7356 

7304  7372  7380  7388  73.90 

LOGARITHMS 


171 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9| 

55 

7404  7412 

7410 

7427 

7435 

7443 

7451  7459 

7466 

7474 

56 

7482  74DO 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559  75(3(3 

7574 

7582 

7589 

7597 

7(504 

7612 

7619 

7627 

58 

7634  |  7(342 

7649 

7657 

7664 

7672 

7(570 

7686 

7094 

7701 

59 

7709  7716 

7723 

7731 

7738 

77  15 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

78 18 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7003 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966  ;  7973 

7980 

7087 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8(135  8041 

8048 

8055 

64 

8002 

800!) 

8075 

8082 

8080 

8096 

8102  ;  8100 

811(5 

8122 

65 

8129 

813(5  8142 

8149 

HI  50 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8209 

8300 

8312 

8319 

68 

8325 

83::  1 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395  |  8401 

8407 

8414 

8420 

842(i 

8432 

8430  8445 

70 

8451 

8457 

84(53 

8470 

8476 

8482 

8488 

S10-J 

8500 

850(3 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8507 

72 

8573 

8579 

8585 

8591 

8597 

800:5 

86(19 

8615 

8621 

8627 

73 

8633 

8639 

8(545 

8651 

8657 

8(563 

8669 

8675 

8(381 

8680 

74 

8692 

8(598 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

87(52 

87(58 

8774 

8779 

8785  1  8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8809 

8904 

8910 1 8915 

78 

8921 

8927 

8032 

8938 

8943 

8949 

8954 

8060 

8965  8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

0000 

0015 

9020  9025 

80 

9031 

903(3 

9042 

9047 

9053 

9058 

9063 

OdOO 

9074 

9079 

81 

9085 

9000 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175  ;  9180 

9186 

83 

9191 

919(5 

9201 

920(3 

0212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9209 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304  9309 

9315 

0320 

9325 

9330 

9335 

0340 

86 

9345 

9350 

0355  03(50 

9365 

0370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405  1  9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455  1  9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504  9509  |  9513 

9518 

9523 

9528 

0533 

9538 

90 

9542 

9547 

9552  9557 

9562 

9566  9571 

9576 

9581  9586 

91 

9590 

9595 

9600  9(505 

9609 

9614 

9010 

9624 

9628 

9(533 

92 

9(338  9643 

9647  9(552 

9657 

9661 

9606 

9671 

9675 

9680 

93 

9685 

9(589  '  9694  i  9(599 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

973(5 

9741 

9745 

9750 

9754 

9759 

9763  9768 

9773 

95 

9777 

9782 

9786 

9791  i  9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836  |  9841 

9845 

9850 

9854 

9859 

9863 

97 

98(38 

9872 

9877 

9881  1  9886 

9890 

9894  0800 

9903 

9008 

98 

9912 

9917 

9921 

9926  9930 

0034 

9939  i  9943 

9948 

0052 

99 

9956 

9961 

9005  \  9969  i  9974 

9978 

9983  ,  9987 

9901  9906 

172  LOGARITHMS 

Find  the  numbers  corresponding  to  the  following  logarithms: 

21.  1.3179.  26.   2.9900.  31.   1.5972.  36.   0.2468. 

22.  3.0146.  27.   0.1731.  32.    1.0011.  37.   0.1357. 

23.  0.7145.  28.   0.8974.  33.    2.7947.  38.    2.0246. 

24.  1.5983.  29.   0.9171.  34.    2.5432.  39.   1.1358. 

25.  2.0013.  30.    3.4015.  35.    0.5987.  40.    4.0478. 

202.  Products  and  powers  may  be  found  by  means  of  loga- 
rithms, as  shown  by  the  following  examples. 

Ex.  1.    Find  the  product  49  x  134  x  .071  x  349. 

Solution.     From  the  table, 

log49  =  l.f>902  or    49  =  lO1-**02. 

log  134  =  2.1271  or  134  =  102-1271. 

log  .071  =2.8513  or  .071  =  102-8618. 

log  349  =  2.5428  or  349  =  102-6428. 

Since  powers  of  the  same  base  are  multiplied  by  adding  exponents, 
§  17G,  we  have       49  x  131  x  .071  x  349  =  10*-2114. 
Hence  log  (49  x  131  x  .071  x  319)  =  5.21 11. 

The  number  corresponding  to  this  logarithm,  as  found  by  the 
method  used  in  Ex.  3  above,  is  162704.  By  actual  multiplication  the 
product  is  found  to  be  162698.914  or  162099  which  is  the  nearest 
approximation  without  decimals.  Hence  the  product  obtained  by 
means  of  logarithms  is  5  too  large.  This  is  an  error  of  about  32^55  of 
the  actual  result  and  is  therefore  so  small  as  to  be  negligible. 

Ex.  2.    Find  (4.05)20. 

Solution.       Log  l.o.l  =  0.0212  or  100-0212  =  1.05. 
Hence  (1.05)20  =  (100-0212)20  =  lO*0-02*2'-20  =  100-424, 

or  log  (1.05)  2°  =  0.4240. 

Hence  (1.05)20  is  the  number  corresponding  to  the  logarithm  0.4240. 


LOGARITHMS  173 

Since  logarithms  are  exponents  of  the  base  10,  it  follows  from 
the  laws  of  exponents  (see  §  198)  that 

(a)  The  logarithm  of  the  product  of  two  or  more  numbers  is 
the  sum  of  the  logarithms  of  the  numbers. 

(b)  The  logarithm  of  a  power  of  a  number  is  the  logarithm  of 
the  number  multiplied  by  the  index  of  the  power. 

That  is, 
log  (a    b  ■  c)  —  log  a  -f  log  b  +  log  c,  and  log  a"  —  n  log  a. 

EXERCISES 

By  means  of  the  logarithms  obtain  the  following  products 
and  powers : 

1.  243  x  76  x  .34.  7.  5.93  x  10.02.  13.  (49)a  x  .19  x  212. 

2.  823.68  x  370.  8.  486  x  3.45.  14.  .21084  x  (.53)2. 

3.  216.83  x  2.03.  9.  (.02)2  x  (0.8).  15.  7.865  x  (.013)2. 

4.  572  x  (.71)2.  10.  (65)2  x  (91)3.  16.  (6.75)3  x  (723)2. 

5.  510x(9.1)3.  11.  (84)2x(75)3.  17.  (1.46)* x  (61.2)". 

6.  43.71  x  (21)2.  12.    (.960)2(49)2.  18.  (3.54)3x  (29.3)2. 

19.    (4.132)2  x  (5.184)2.  20.    1946  x  398  x  .08. 

203.   Quotients  and  roots  may  be  found  by  means  of  logarithms, 
as  shown  by  the  following  examples. 

Ex.  1.    Divide  379  by  793. 
Solution.     From  the  table, 

log  379  =  2.5786  or  102-6™5  =  379. 

log  793  =  2.8993  or  102-^  _  793. 

Hence  by  the  law  of  exponents  for  division,  §  175, 

379  +  793  =   102-67S6-2.8993. 

Since  in  all  operations  with  logarithms  the  mantissa  is  positive,  write 
the  first  exponent  3.5786  -  1  and  then  subtract  2.8993. 

Hence  log  (379  -  793)  =.6793  -  1  =  1.6793. 

Hence  the  quotient  is  the  number  corresponding  to  this  logarithm. 


174  LOGARITHMS 


Ex.  2.    By  means  of  logarithms  approximate  V  42-  x  375. 
By  the  methods  used  above  we  find 

log(422  x  375)  =  11.0874  or  10U<>8H  =  42-  x  375. 
Hence  V422  x  375  =  (lO11-0874)?  =  10    3     =  lO3®58. 


That  is,  log  V422  x  375  =  3.6958. 

Hence   the   result    sought   is  the    number   corresponding   to   this 
logarithm. 

It  follows  from  the  laws  of  exponents  (see  §  198)  that 

(a)  TJie  logarithm   of  a  quotient  equals  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor. 

(b)  TJie  logarithm  of  a  root  of  a  number  is  the  logarithm  of 
the  number  divided  by  the  index  of  the  root. 

That  is 

log  -  =  log  a  —  log  b  and  log  V  a  =  ^—-  • 
b  n 

EXERCISES 

By  means  of   logarithms    approximate  the  following  quo- 
tients and  roots : 

1.  45.2-5-8.9.         4.  V196  x  25G.  7.  Vl5  x  ^67. 

2.  231.18  +  42.      5.   5334  x. 02374,  ^2Ux^34T. 

27.43x3.246 

3.  .04Q05  +  . 327.    6.   </69-s--^21.  9-  (5184)*+ (38124)*. 


10.  (6.75)3  +  (2.132)2.  s/  13*  x  .31'  x  4.313' 

lb.  \j —  — ==. 

11.  #105  +  ^76.  vV71xv41xV51 

12.  (91125)* +  (576)*.  ,!      4,  x  >r>73  x  4l>a 

17.  \/-rr= 

13.  (3.040)3  -  (.0005)3.  x  s!/o.2  x  V.83  x  V23 

14.  (29.3)Uv(3,i:1.       ig  /  mx<mx</7\ 

15.  S^39  x  </W  x  \  87.  W47  x  a/71  x  (.003)* 


CHAPTER   XII 
PROGRESSIONS 

ARITHMETIC    PROGRESSIONS 

204.  An  arithmetic  progression  is  a  series  of  numbers,  such 
that  any  one  after  the  first  is  obtained  by  adding  a  fixed  num- 
ber to  the  preceding.  The  fixed  number  is  called  the  common 
difference. 

The  general  form  of  an  arithmetic  progression  is 

a,   a  +  d,   a  +  2d,    a  +  3d,    •  ■•, 

where  a  is  the  first  term  and  d  the  common  difference. 

E.g.  2,  5,  8,  11,  14,  •••  is  an  arithmetic  progression  in  which  2  is  the 
first  term  and  3  the  common  difference.  Written  in  the  general  form, 
itwouldbe      ^   2  +  3j   2  +  2.3)   0  +  3-3,   2+4-3,    .... 

205.  If  there  are  n  terms  in  the  progression,  then  the  last 
term  is  a-\-(n  —  l)d.     Indicating  the  last  term  by  I,  we  have 

/=:a+(n-l)d.  I 

An  arithmetic  progression  of  n  terms  would  then  be  written 
in  general  form,  thus, 

a,   a  +  d,   a  +  2d,    ••  ,    a+(n-2)d,   a+(n  —  l)d. 

EXERCISES 

1.    Solve  I  for  each  letter  in  terms  of  all  the  others. 

In  each  of  the  following  find  the  value  of  the  letter  not  given,  and 
write  out  the  progression  in  each  case. 


a  =  2, 

fa  =  3, 

fo  =  l, 

(a=7, 

d  =  2, 

3. 

\d  =  5, 

4. 

11  =  15, 

5. 

\n  =  :n, 

n  =  7. 

{ I  =  43. 

I  =  15. 

U  =  91. 

175 


176  PROGRESSIONS 


a  =  4,  (a  =3,  rd  =  —  5,  fo  =  ll, 

d  =  — 3,      8.    |d  =  — 5,    10.   |n  =  13,       12.      Z  =  — 39, 


n  =  18.  I  —  -  32.  /  =  -  63.  \d  =  -5 


a  =3, 

[d  =  -5, 

d  =  — 

5, 

10. 

a  =  13, 

Z  =  - 

32. 

I  /  =  -  63. 

c?  =  7, 

fo  =  -3, 
«  =  9, 
U  =  -27. 

w  =  8, 

11. 

2  =  24 

a  =  x, 
d  =  i,         9.    |n  =  8,        11.    J7i  =  9,         13.    ll  =  y, 

n  =  7.  [1  =  24:.  [z  =  -27.  U  =  «- 

206.  The  sum  of  an  arithmetic  progression  of  n  terms  may  be 
obtained  as  follows : 

Let  sn  denote  the  sum  of  n  terms  of  the  progression.     Then, 

s„  =  a  +  [a  +  d]  +  [a  +  2  rf]  +  ...  +  [a  +  («  -  2>/]  +  [a  +  (n  -  l)r/].    (1) 

This  may  also  be  written,  reversing  the  order  of  the  terms,  thus, 

s„  =  [a  +  (»  -  1>/]  +  [a  +(n - 2)d]  +  ••■  +  [a  +  2 d]  +  [a  +  d]  +  a.  (2) 

Adding  (1)  and  ('2),  we  have 
2  s„  =  [2  a  +  (»  -  1)'/]  +  [2  a  +  (n  -  2)d  +  d] 

+  ...  +  [2  a  +  („  -  2)d  +  d]  +  [2  a  +  (n  -  l)d]. 

The  expression    in   each    bracket   is    reducible   to  '2  n  +  (>i  —  l)r/, 
which  may  also  be  written  a  +  [«  +  (»  —  IV]  =  a  +  I,  by  §  205. 
Since  there  are  n  of  these  expressions,  each  a  +  I,  we  have 

2sn  =  n(a  +  I). 

Hence  s„  =  -  (a  +  /).  II 

& 

This  formula  for  the  sum  of  n  terms  involves  a,  I,  and  n,  that 
is,  the  first  term,  the  last  term,  and  the  number  of  terms. 

207.  In  the  two  equations, 

l  =  a+(n-l)d,  I 

.-5(0  +  1),  II 

there  are  live  letters,  namely,  a,  d,  I,  n,  s.  If  any  three  of 
these  are  given,  the  equations  I  and  II  may  be  solved  simul- 
taneously to  find  the  other  two,  considered  as  the  unknowns. 


ARITHMETIC  PROGRESSIONS  177 

The  solution  of  problems  in  arithmetic  progression  by  means 
of  equations  I  and  II  is  illustrated  in  the  following  examples: 

Ex.  1.    Given  n  =  ll,   1  =  23,   s  =  143.     Find  a  and  d. 
Substituting  the  given  values  in  I  and  II, 

23  =  a+(ll-l)d.  (1) 

U3  =  V(«  +  23).  (2) 

From  (2),  a  =  3,  which  in  (1)  gives  d  =  2. 

Ex.  2.    Given  d  =  4,   n  =  5,   s  =  75.     Find  a  and  ?. 
From  I  and  II,  I  =  a  +  (5  -  1)4,  (1) 

75=|(a  +  Q.  (2) 

Solving  (1)  and  (2)  simultaneously,  we  have  a  =  7,   /  =  23. 

Ex.3.    Given  rf  =  4,    1  =  35,   s  =  1G1.     Find  a  and  n. 
From  I  and  II,  35  =  a  +  (n  -  1)4,  (1) 

161=|(a  +  35).  (2) 

From  (1)  a  =  39  -  in, 

which  in  (2)  gives  161  =  -(74  -  4  n)  =  37  »  -  2  n2. 

Hence  n  =  -223,  or  7. 

Since  an  arithmetic  progression  must  have  an  integral  number  of 
terms,  only  the  second  value  is  applicable  to  this  problem. 

Ex.  4.    Given  d  =  2,    I  =  11,    s  =  35.     Find  a  and  n. 
Substituting  in  T  and  U,  and  solving  for  a  and  n,  we  have 
a  =  3,    n  —  5,    and   a  =  —  1,    n  =  7. 

Hence  there  are  two  progressions, 

-  1,  1,  3,  5,  7,  0,  11, 
and  3,  5,  7,  0,  11, 

each  of  which  satisfies  the  given  conditions. 


178  PROGRESSIONS 


EXERCISES 

In  each  of  the  following  obtain  the  values  of  the  two  letters 
not  given. 

If  fractional  or  negative  values  of  n  are  obtained,  such  a  result  in- 
dicates that  the  problem  is  impossible.  This  is  also  the  case  if  an 
imaginary  value  is  obtained  for  any  letter.  In  each  exercise  interpret 
all  the  values  found. 

rs  =  96,  f.s  =  88,  \d=-l,  [d=6, 

1.  ?  =  19,       4.       /=-7,      7.    |w=4l,         10.    j  2=49, 

U=2.  U=-3.  [l=-3o.  [s=232. 

rs=34,  fn=18,  (1  =  30,  U=7, 

2.  <  =  14,       5.      a=4,  8.      s=162,       11.    \d=n, 

|r/  =  7,  \n=U,  (a  =30,  rs=14, 

3.  I  1  =  27,         6.     ja=7,  9.        »  =  10,  12.       r/=:3, 
[s  =  187.            U=14.               [s=120.  [z  =  4. 

In  each  of  the  following  call  the  two  letters  specified  the  unknovms 
and  solve  for  their  values  in  terms  of  the  remaining  three  letters 
supposed  to  be  known. 

13.  a.  <i.  15.    a,  n.  17.    d,  I.  19.    <!,  s.  21.    /,  s. 

14.  a,  I  16.    a,  s.  18.    <l.  n.        20.    /.  n.  22.   ?j,  s. 

208.  Arithmetic  means.  The  terms  between  the  first  and  Hie 
last  of  an  arithmetic  progression  are  called  arithmetic  means. 

Thus,  in  2,  .">,  8,  11,  11,  17,  the  four  arithmetic  means  between  2 
and  17  are  ."J,  8,  11,  11. 

If  the  first  and  the  last  terms  and  the  number  of  arith- 
metic means  between  them  are  given,  then  these  means  can  be 
found. 

For  we  have  given  a,  I,  and  n.  Hence  <1  can  lie  found  and  the 
whole  series  constructed. 


ARITHMETIC  PROGRESSIONS  179 

Example.     Insert  7  arithmetic  means  between  3  and  19. 

In  this  progression  a  =  3,   /  =  19,  and  n  =  9. 

Hence,  from  1  =  a  +  (n  —  l)d  we  find  d  =  2  and  the  required  means 
are  5,  7,  9,  11,  13,  15,  17. 

209.  The  case  of  one  arithmetic  mean  is  important.  Let  A 
be  the  arithmetic  mean  between  a  and  I.  Since  a,  A,  I  are  in 
arithmetic    progression,    we    have   A  =  a  4-  d,    and    1  =  A  +  d. 

Hence  A-l  =  a-A 

A^"^1-  HI 

EXERCISES  AND  PROBLEMS 

1.  Insert  5  arithmetic  means  between  5  and  —  7. 

2.  Insert  3  arithmetic  means  between  —  2  and  12. 

3.  Insert  8  arithmetic  means  between  —  3  and  —  5. 

4.  Insert  5  arithmetic  means  between  —  11  and  40. 

5.  Insert  15  arithmetic  means  between  1  and  2. 

6.  Insert  9  arithmetic  means  between  2|  and  —  11. 

7.  Find  the  arithmetic  mean  between  3  and  17. 

8.  Find  the  arithmetic  mean  between  —  4  and  16. 

9.  Find   the   tenth    and   eighteenth    terms    of   the   series 
4,  7,  10,  -. 

10.  Find  the  fifteenth  and  twentieth  terms  of  the  series 
-8,  -4,0,.-.. 

11.  The  fifth  term  of  an  arithmetic  progression  is  13  and 
the  thirtieth  term  is  49.     Find  the  common  difference. 

12.  Find  the  sum  of  all  the  integers  from  1  to  100. 

13.  Find  the  sum  of  all  the  odd  integers  between  0  and  200. 

14.  Find  the  sum  of  all  integers  divisible  by  6  between  1 
and  500. 

15.  Show  that  1  4-  3  +  5  +  •••  +  n  =  n2  where  n  is  any  odd 
integer. 


180  PBOGEESSIONS 

16.  In  a  potato  race  40  potatoes  are  placed  in  a  straight  line 
one  yard  apart,  the  first  potato  being  two  yards  from  the 
basket.  How  far  must  a  contestant  travel  in  bringing  them 
to  the  basket  one  at  a  time  ? 

17.  There  are  three  numbers  in  arithmetic  progression  whose 
sum  is  15.  The  product  of  the  first  and  last  is  31-  times  the 
second.     Find  the  numbers. 

18.  There  are  four  numbers  in  arithmetic  progression  whose 
sum  is  20  and  the  sum  of  whose  squares  is  120.  Find  the 
numbers. 

19.  If  a  body  falls  from  rest  16.08  feet  the  first  second, 
48.24  feet  the  second  second,  80.40  the  third,  etc.,  how  far  will 
it  fall  in  10  seconds  ?  15  seconds  ?  t  seconds  ? 

20.  According  to  the  law  indicated  in  problem  19  in  how 
many  seconds  will  a  body  fall  1000  feet  ?  s  feet  ? 

If  a  body  is  thrown  downward  with  a  velocity  of  v0  feet  per  second, 
then  the  distance,  s,  it  will  fall  in  t  seconds  is  v0t  feet  plus  the  distance 
it  would  fall  if  starting  from  rest. 

That  is,  5  =  VQt  +  \ gf,  where  g  =  32.10. 

21.  In  what  time  will  a  body  fall  1000  feet  if  thrown  down- 
ward with  a  velocity  of  20  feet  per  second  ? 

22.  "With  what  velocity  must  a  body  be  thrown  downward 
in  order  that  it  shall  fall  360  feet  in  3  seconds  ? 

23.  A  stone  is  dropped  into  a  well,  and  the  sound  of  its 
striking  the  bottom  is  heard  in  3  seconds.  How  deep  is  the 
well  if  sound  travels  1080  feet  per  second? 

A  body  thrown  upward  with  a  certain  velocity  will  rise  as  far  as  it 
would  have  to  fall  to  acquire  this  velocity.  The  velocity  (neglecting 
the  resistance  of  the  atmosphere)  of  a  body  starting  from  rest  is  gt 
where  g  =  32.16  and  /  is  the  number  of  seconds. 

24.  A  rifle  bullet  is  shot  directly  upward  with  a  velocity  of 
2000  feet  per  second.  How  high  will  it  rise,  and  how  long  be- 
fore it  will  reach  the  ground  '.' 


GEOMETRIC  PROGRESSIONS  181 

25.  From  a  balloon  5800  feet  above  the  earth,  a  body  is 
thrown  downward  with  a  velocity  of  40  feet  per  second.  In 
how  many  seconds  will  it  reach  the  ground  ? 

26.  If  in  Problem  25  the  body  is  thrown  upward  at  the  rate 
of  40  feet  per  second,  how  long  before  it  will  reach  the  ground  ? 

GEOMETRIC    PROGRESSIONS 

210.  A  geometric  progression  is  a  series  of  numbers  in  which 
any  term  after  the  first  is  obtained  by  multiplying  the  pre- 
ceding term  by  a  fixed  number,  called  the  common  ratio. 

The  general  form  of  a  geometric  progression  is 

a,   an,   ar,   ar\  •••,  ar"\ 

in  which  a  is  the  first  term,  r  the  constant  multiplier,  or  com- 
mon ratio,  and  n  the  number  of  terms. 

E.g.  3,  6,  12,  24,  48,  is  a  geometric  progression  in  which  3  is  the 
first  term,  2  is  the  common  ratio,  and  5  is  the  number  of  terms. 

Written  in  the  general  form  it  would  be  3,  3  •  2,  3  •  2'2,  3  •  23,  3  •  24. 

211.  If  I  is  the  last  or  nth  term  of  the  series,  then 

/  =  ar"~\  I 

If  any  three  of  the  four  letters  in  I  are  given,  the  remaining 
one  may  be  found  by  solving  this  equation. 

EXERCISES 

In  each  of  the  following  find  the  value  of  the  letter  not  given  : 


4.       r  =  -2,       7.       r  =  -3  10. 


11. 


3.       r  =  -3,     6.       r  =  2,  9.       r  =  -f,      12 


a  =  —  1, 

f  a  =  —  \, 

r  =  -2, 

7. 

\  r  =  h 

n  =  9. 

[n  =  0. 

1  =  1024, 

U  =  6. 

r  =  -2, 

8. 

n  =  11. 

1  =  1024, 

r  z=-i6 

U- 1. 

[n  =  S. 

r  =  2, 

9. 

>i  =  ll. 

182  PROGRESSIONS 

212.    The  sum  of  n  terms  of  a  geometric  expression  may  be 
found  as  follows : 

If  s„  denotes  the  sum  of  n  terms,  then 

s„  =  a  +  ar  +  ar-  +  ■■■  +  ar'1-2  +  arn~l.  (1) 

Multiplying  both  members  of  (1)  by  r,  we  have 

rsn  =  ar  +  ar'-  +  ar3  +  •••  +  ar'-1  +  ar".  (2) 

Subtracting  (1)  from  (2).  and  canceling  terms,  we  have 


?>•„  —  s„  =  ar™  —  a. 


(3) 


Solving  (3)  for  sn  we  have 


_ar"-a_a(r"-l) 
r— 1  r  — 1 

This  formula  for  the  sum  of  w  terms  of  a  geometric  series 
involves  only  a,  r,  and.  n. 

Since  ar"  =  r  •  arnl  =  r  •  /,  s"  may  also  be  written  : 

^  =  W-a     o-W.  m 

r  —  1       1  —  r 

This  formula  involves  only  r,  I,  and  a. 

213.  From  equations  I  and  II  or  I  and  III  any  two  of  the 
numbers  a,  I,  r,  s,  and  n  can  be  found  when  the  other  three  are 
given,  as  in  the  following  examples. 

Ex.  1.    Given  n  =  7,  r  =  2,  s  =  381.    Find  a  and  I. 

From  I  and  III,  Z  =  a-2fi=6ia,  (1) 

381  =  y~~  =  21 -a.  (2) 

Substituting  /  =  61a  in  (2),  we  obtain  a  —  3,  and  /  =  192. 

Ex.  2.    Given  a  =  -  3,  I  =  -  243,  s  =  -  183.    Find  r  and  ». 
From  I  and  III.  -  213  =  (-  Sy»~\  (1) 

_183  =  -243r  +  3,  (L)) 

r  —  1 


GEOMETRIC  PROGRESSIONS 


183 


From  (2)  r  =  -  3.  (3) 

From  (1)  81  =  (-3)*"1.  (4) 

Since  (  —  3)4  =  81,  we  have  n  —  1  =  4  or  n  =  5. 

EXERCISES 

1.  Solve  II  for  a  in  terms  of  the  remaining  letters. 

2.  Solve  III  for  each  letter  in  terms  of  the  remaining  letters. 
In  each  of  the  following  find  the  terms  represented  by  the 

interrogation  point. 


3. 


a  =  1, 
r  =  3, 

n  =  5, 
s  =  ? 

a 

S  =  f  ¥5 


s  =  635, 
r  =  2, 
n  =  7, 
a  =  ? 


n  = 
a  — 


13, 

f» 

? 


1- 


?i  =  5, 
£  =  1296,   9. 


—  ? 


8  =  1050 1,   10. 

Z=? 


a  = 


9 

5, 

2 
"3* 

9. 
2  J 

=  7, 


1  6. 

8l> 


81? 


214.  Geometric  means.  The  terms  between  the  first  and  the 
last  of  a  geometric  progression  are  called  geometric  means. 

Thus  in  3,  6,  12,  24,  48,  three  geometric  means  between  3  and  48 
are  6,  12  and  24. 

If  the  first  term,  the  last  term,  and  the  number  of  geometric 
means  are  given,  the  ratio  may  be  found  from  I,  and  then  the 
means  may  be  inserted. 

Example.    Insert  4  geometric  means  between  2  and  64. 
We  have  given  a  =  2,1  =  64,  n  =  4  +  2  =  6.  to  find  r. 
From  [,  64  =  2  •  r6  -1  or  r5  =  32  and  r  =  2. 

Hence,  the  series  is  2,  4,  8,  16,  32,  64. 

215.  The  case  of  one  geometric  mean  is  important.     If  G  is 

the  geometric  mean  between  a  and  b,  we  have  —  =  — 

a      G 


Hence, 


Va6. 


184  PROGRESSIONS 

216.  Problem.  In  attempting  to  reduce  §  to  a  decimal,  Ave 
find  by  division  .006  •••,  the  dots  indicating  that  the  process 
goes  on  indefinitely. 

Conversely,  we  see  that  .006  •  ■  •  =  T%  +  jfa  +  y^-  +  •  • .,  that 
is,  a  geometric  progression  in  which  a  =  T6TF,  r  =  T1jJ,  and  n  is 
not  fixed  but  goes  on  increasing  indefinitely. 

As  n  grows  large,  /  grows  small,  and  by  taking  n  sufficiently  large, 
I  can  be  made  as  small  as  we  please.  Hence  formula  III,  §  I'll',  is  to 
be  interpreted  in  this  case  as  follows : 

a  -  rl      10      10      6  -  I 


1  -  /•         Y  _  J_  9 

10 

in  which  /  grows  small  indefinitely  as  n  increases  indefinitely,  so  that 
by  taking  n  large  enough  sn  can  be  made  to  differ  as  little  as  we  please 

from  =  -  =     • 

9         9      3 

In  this  case  we  say  sH  approaches  las  a  limit  as  n  increases 
indefinitely. 

Observe  that  this  interpretation  can  apply  only  when  the 
constant  multiplier  r  is  a  proper  fraction. 

EXERCISES   AND   PROBLEMS 

1.  Insert  5  geometric  means  between  2  and  128. 

2.  Insert  7  geometric  means  between  1  and  7i^. 

3.  Find  the  geometric  mean  between  8  and  18. 

4.  Find  the  geometric  mean  between  TV  and  \. 

5.  Find  the  fraction  which  is  the  limit  of  .333  •••. 

6.  Find  the  fraction  which  is  the  limit  of  .1666  ■•-. 

7.  Find  the  fraction  which  is  the  limit  of  .08333  •••. 

8.  Find  the  13th  term  of  - -1/,  4,  -3  •••. 

9.  Find  the  sum  of  15  terms  of  the  series  —  243,  81,  —  L'7  •  •  •. 
10.  Find  the   limit  of  the   sum    |  +  |  +  |  +  £+  ...,  as  the 

number  of  terms  increases  indefinitely. 


GEOMETRIC  PROGRESSIONS  185 


Given 

Find 

G 

iven 

Find 

11.   a,  r, 

n 

l,s 

15. 

a,  n,  I 

s,  r 

12.    a,  r, 

s 

I 

16. 

r,  n,  I 

s,  a 

13.    r,  n, 

s 

I,  a 

17. 

r,  I,  s 

a 

14.    a,  r, 

I 

s 

18. 

a,  I,  s 

r 

19.  The  product  of  three  terms  of  a  geometric  progression 
is  1000.     Find  the  second  term. 

20.  Four  numbers  are  in  geometric  progression.  The  sum 
of  the  second  and  third  is  18,  and  the  sum  of  the  first  and 
fourth  is  27.     Find  the  numbers. 

21.  Find  an  arithmetic  progression  whose  first  term  is  1 
and  whose  first,  second,  fifth,  and  fourteenth  terms  are  in  geo- 
metric progression. 

22.  Three  numbers  whose  sum  is  27  are  in  arithmetic  pro- 
gression. If  1  is  added  to  the  first,  3  to  the  second,  and  11  to 
the  third  the  sums  will  be  in  geometric  progression.  Find 
the  numbers. 

23.  To  find  the  compound  interest  when  the  principal,  the 
rate  of  interest,  and  the  time  are  given. 

Solution.  Let  p  equal  the  number  of  dollars  invested,  r  the  rate  of 
per  cent  of  interest,  t  the  number  of  years,  and  a  the  amount  at  the 
end  of  t  years. 

Then      a  =  p(l  +  r)  at  the  end  of  one  year. 

a  —  p  ( 1  +  r)  (1  +  r)  =  p  (1  +  r)a  at  the  end  of  two  years. 

and  a  —  p(l  +  r)'  at  the  end  of  t  years. 

That  is,  the  amount  for  t  years  is  the  last  term  of  a  geometric  pro- 
gression in  which  p  is  the  first  term,  1  +  r  is  the  ratio,  and  t  +  1  is  the 
number  of  terms. 

24.  Show  how  to  modify  the  solution  given  under  problem 
23  when  the  interest  is  compounded  semiannually;  quarterly. 

25.  Solve  the  equation  a=p(\.  +  r)'  forp  and  for  r. 


186  PROGRESSIONS 

26.  Solve  a  =p(l  +  '">'  f°r  t. 

Solution,     log  a  =  logp(l  +  /•)'  =  log/)  4-  log  (1  +  r)1 

=  log  p  +  /  loo-  (1  +  r).    (See  §  202.)     Hence  i  =  loga~log;j- 

log  (1  +  r) 

27.  At  what  rate  of  interest  compounded  annually  will 
$1200  amount  to  $1800  in  12  years'.' 

28.  At  what  rate  of  interest  compounded  semiannually  will 
a  sum  double  itself  in  20  years  ?  in  15  years  ?  in  10  years  ? 

29.  In  what  time  will  $8000  amount  to  $13,500,  the  rate 
of  interest  being  3^  %  compounded  annually  ? 

30.  In  what  time  will  a  sum  double  itself  at  3  %,  4  %,  5  ' ,  . 
compounded  semiannually  '.' 

The  present  value  of  a  debt  due  at  some  future  time  is  a  sum  such 
that,  if  in  vested  at  compound  interest,  the  amount  at  the  end  of  the 
time  will  equal  the  debt. 

31.  What  is  the  present  value  of  $2500  due  in  4  years, 
money  being  worth  3.1,  >J0  interest  compounded  semiannually? 

32.  A  man  bequeathed  $50,000  to  his  daughter,  payable  on 
her  twenty-fifth  birthday,  with  the  provision  that  the  present 
worth  of  the  bequest  should  be  paid  in  case  she  married  before 
that  time.  If  she  married  at  21,  howmuch  would  she  receive, 
interest  being  A'/(l  per  annum  and  compounded  quarterly? 

33.  What  is  the  rate  of  interest  if  the  present  worth  of 
si' 1.000  due  in  7  years  is  $19,500? 

34.  In  how  many  years  is  $5000  due  if  its  present  worth 
is  $3500,  the  rate  of  interest  being  3|  <f0  compounded  annually? 

HARMONIC    PROGRESSIONS 
217.    A  harmonic   progression    is    a    series    whose   terms    arc 
the  reciprocals  of  the  corresponding  terms  of  an   arithmetic 
progression. 

E.g.     1.    .    .    .      •••  is  a  harmonic  progression  whose  terms  are  the 

:i  .">  7  it 
reciprocals  of  the  terms  of  the  arithmetic  progression  1.  ">.  5,  7.  '.)  ■•-. 


HARMONIC  PROGRESSIONS  187 

The  name  harmonic  is  given  to  such  a  series  because  musical  strings 
of  uniform  size  and  tension,  whose  lengths  are  the  reciprocals  of  the 

positive  integers,  i.e.  1,  -,  -,  -  •••,  vibrate  in  harmony. 
2  3   4 

The  general  form  of  the  harmonic  progression  is 

1    _1 i_  1  j 

a'  a  +  </'a  +  2</'  '"'    a  +  (n-l)d 

It  follows  that  if  a,  b,  c,  d,  e,  ...  are  in  harmonic  progression, 

then-,-,-,-,-,   •••  are  in  arithmetic  progression.     Hence,  all 
a  bode 

questions  pertaining  to  a  harmonic  progression  are  best   an- 
swered by  first  converting  it  into  an  arithmetic  progression. 

218.  Harmonic  means.  The  terms  between  the  first  and  the 
last  of  a  harmonic  progression  are  called  harmonic  means  be- 
tween them. 

Example.     Insert  five  harmonic  means  between  30  and  3. 

This  is  done  by  inserting  five  arithmetic  means  between  -^l  and  i. 
By  the  method  of  §  20S  the  arithmetic  series  is  found  to  be  ^,  r\,  T25, 
iff)  37o'  Jo'  h     Hence,  the  harmonic  series  is  30,  12,  -^,  f£,  -3^,  f  S,  3. 

219.  The   case    of   a  single   harmonic   mean   is   important. 

Ill 
Let  a,  H,  I  be  in  harmonic  progression.     Then  -,  — ,  -  are  in 

arithmetic  progression.  a 

M 

Hence,  by  §  209,  i  =  ?-_i  or  H  =  f^-/- 

220.  The  arithmetic,  geometric,  and  harmonic  means  be- 
tween a  and  I  are  related  as  follows : 


We  have  seen      A  =  a       ,  G  =  vV,  //  =  — — -. 
2  a  +  I 

Hence,  A  =  «±i  ^  al  =  «±i. 

G1         2  2  al 

Therefore,  —  =  -,  or  A  =  -. 

G*      H        G      II 

That  is,  G  is  a  mean  proportional  between  .1  and  //.     See  §  172. 


188  PROGRESSIOXS 

EXERCISES   AND   PROBLEMS 

1.  Insert  three  harmonic  means  between  22  and  11. 

2.  Insert  six  harmonic  means  between  \  and  -2#. 

3.  The  first  term  of  a  harmonic  progression  is  i  and  the 
tenth  term  is  ^V     Find  the  intervening  terms. 

4.  Two  consecutive  terms  of  a  harmonic  progression  are  5 
and  6.  Find  the  next  two  terms  and  also  the  two  preceding 
terms. 

5.  If  a,  b,  c  are  in  harmonic  progression,  show  that 
a  -=-  c  =  (a  —  b)^-(b  —  c). 

6.  Find   the    arithmetic,  geometric,  and   harmonic   means 

In 'tween : 

(a)  16  and  36 ;  (b)  m  +  n  and  m  —  n  ;  (c) and 


m  +  n  m  —  n 

7.  The  harmonic  mean  between  two  numbers  exceeds  their 
arithmetic  mean  by  7,  and  one  number  is  three  times  the  other. 

Find  the  numbers. 

8.  If  x,  y,  and  z  are  in  arithmetic  progression,  show  that 
nix,  my,  and  mz  are  also  in  arithmetic  progression. 

9.  x,  y,  and  z  being  in  harmonic   progression,  show   that 

x  1/  ,  z  .  . 

-,  and  —  ate  m  harmonic  progression, 


x  +  y  +  z  x  +  y  +  z  x  +  y  +  z 

and  also  that  —  —  — •—,  and are  in  harmonic  progression. 

y  +  z  x  +  z  .'•  +  .7 

10.  The  sum  of  three  numbers  in  harmonic  progression  is  3, 
and  the  first  is  double  the  third.     Find  the  numbers. 

11.  The  geometric  mean  between  two  numbers  is  \  and  the 
harmonic  mean  is  !.     Find  the  numbers. 

12.  Insert  n  harmonic  means  between  the  numbers  a  ami  //. 


CHAPTER    XIII 

THE   BINOMIAL   FORMULA 

221.    In  Chapter  II  the  following  products  were  obtained : 
(«  +  6)2  -  a2  +  -2  ah  +  ft2. 
(a  +  b)8  =  a3  +  3  a2b  +  3  air  +  b3. 
(a  +  hy  =  a4  +  i  asb  +  6  a2b*  +  1  ab3  +  6*. 
(a  +  b)5  =  «5  +  5  a4!*  +  10  a862  +  10  a?b3  +  5  ^A4  +  b5. 

By  a  study  of  these  the  following  facts  may  be  observed : 

1.  Each  product  has  one  term  more  than  the  number  of  units  in 
the  exponent  of  the  binomial. 

2.  The  exponent  of  a  in  the  Jirst  term  is  the  same  as  the  exponent 
of  the  binomial,  and  diminishes  by  unity  in  each  succeeding  term. 

The  exponent  of  b  in  the  last  term  is  the  same  as  the  exponent  of 
the  binomial,  and  diminishes  by  unity  in  each  preceding  term. 

3.  The  sum  of  the  exponents  in  each  term  is  equal  to  the  expo- 
nent of  the  binomial. 

4.  The  coefficient  of  the  first  term  is  unity;  of  the  second  term, 
the  same  as  the  exponent  of  the  binomial ;  and  the  coefficient  of 
any  other  term  may  be  found  by  multiplying  the  coefficient 'of  the 
next  preceding  term  by  the  exponent  of  a  in  that  term  and  dividing 
this  product  by  a  number  one  greater  than  the  exponent  of  b  in 
that  term. 

5.  The  coefficients  of  any  pair  of  terms  equally  distant  from  the 
ends  are  equal. 

Statements  2  and  4  form  a  rule  for  writing  out  any  power 
of  a  binomial  up  to  the  fifth.     Let  us  find  (a  +  b)6. 

189 


190  THE   BINOMIAL    FORMULA 

Multiplying  (a  +  b)5  by  a  +  b,  we  have 

(a  +  b)5(a  +  6)  =  a6  +  5  a5i  +  10  a4*2  +  10  o8A3  +    5  a2i4  +    <//<5 
r/5/;+    5a4fe-  +  10  rrV,*8  +  10  aHA  +  5  ab5  +  b6 

Hence  (a  +  b)6  =  a6  +  6  «56  +  15  rt4//2  +  20  a%3  +  15  a2b*  +  6  ab5  +V6 
From  this  it  is  seen  that  the  rule  holds  also  for  (a  +  b)6. 

PROOF    BY   MATHEMATICAL    INDUCTION 
222.    A    proof    that    the    above    rule    holds    for    nil  positive 

integral  powers  of  a  binomial  may  lie  made  as  follows: 

First  step.     Write  out  the  product  as  it  would  be  for  the  nth 

power  on  the  supposition  that  the  rule  holds. 

Then  the  firs!  term  would  be  an  ami  the  lust  term  b".     The  second 

terms  from  the  ends  would  lie  nan~1b  and  nabn~l.     The  third  terms 

i-  ,i  t  ,11       n(n  —  1)     „   ojo         i   n(n  —  1)    .>,„   0      rrl 

i  roni   the  ends  would   be  — ^ —  — -  an~2b2  and  — ^ —    --'  <i -/>»--.      I  lie 

1-2  1-2 

fourth  terms  from  the  ends  would  be 

u(n  —  l)(/i  -  2)     _    .,,„         ,  h(i)  -  \)(n  —  2)     „,„_» 

— y-v —  — ^  an~W  and  — * —  — ^ -  asb"  ■: 

1-2-3  1-2.3 

and  so  on,  giving  by  the  hypothesis, 

(„  1  /,)•>  =  a" +  >«i<>-ib  +  ^~±)<vi--b~+  ■■■  +l^!l^ln-b»-2+tiob"    i     /.". 
1-2  1  •  2 

Second  step.     Multiply  this  expression  by  a  +  b  and   see  if 

the  result  can  be  so  arranged  as  to  conform  to  the  same  rule. 

Then,  (a  +  b)"(a  +  b) 

, .            .   ,  >t(»  -  })       ,,„  „.  ,        , 

=  «" fl  +  na"b  +  -±-— -a"-lb-  +  ■  ■  ■  +  na-b"-1  +  ab" 

a"h  +  )in"-Vi-  +  •  •  •  +  — k- — —  a26n_1  +  «a6n  +  6n+1. 

Hence  adding, 
(a  +  b)n+1  =  an+1  +  (n  +  l)a"b  +  [~"v"  ~  ^  +  n~|  a""1*)2  +  •  •  • 

+  T„  +  El"  ~  ^"l  a26n-l  +  („  +  l)ajn  +  &»+l. 

Combining  the  terms  in  brackets,we  have, 

(a  +  b)n+i  =  a"+l  +  (n  +  })>,»/>  +  ("   "  }>)", /»-'//-  +  •  •  ■ 


PROOF  BY  INDUCTION  191 

The  last  result  shows  that  the  rule  holds  for  (a  +  b)n  +  1  if  it 
holds  for  (a  +  by.  That  is,  if  the  rule  holds  for  any  positive  inte- 
gral, exponent,  it  holds  for  (he  next  higher  integer. 

Third  step.  It  was  found  above  by  actual  multiplication  that 
the  rule  does  hold  lor  (a  +  bf.  Hence  by  the  above  argument 
we  know  that  the  rule  holds  for  (a  +  by. 

Moreover,  since  we  now  know  that  the  rule  holds  for  (a  +  l>)~,  we 
conclude  by  the  same  argument  that  it  holds  for  (a  +  6)8,  and  if  for 
(u  +  b)s,  then  for  (a  +  b)9,  and  so  on. 

Since  this  process  of  extending  to  higher  powers  can  be  car- 
ried on  indefinitely,  we  conclude  that  the  live  statements  in 
§221  hold  for  all  positive  integral  powers  of  a  binomial. 

The  essence  of  this  proof  by  mathematical  induction  consists 
in  applying  the  supposed  rule  to  the  nth  power  and  finding 
that  the  rule,  does  hold  for  the  (//  -fl)th  power  if  it  holds  for 
the  nth  power. 

223.  The  general  term.  According  to  the  rule  now  known  to 
In ild  for  any  positive  integral  exponent,  we  may  write  as  many 
terms  of  the  expansion  of  {a  +  &)"  as  may  be  desired,  thus : 

(a  +  b)"  =  a"  +  na"    '6  +  /7^=-^a»   -6- 

1  •  2 

„(„_1)(„_2)  /i(fi-l)(/i-2)(ii-3)      4ft4  ■  ...     i 

^        1.2- 3  1-  2-3-4 

From  this  result,  called  the  binomial  formula,  we  see : 

(1)  The  exponent  of  b  in  any  term  is  one  less  than  the  number  of 
that  term,  and  the  exponent  of  a  is  n  minus  the  exponent  of  b.  Hence 
the  exponent  of  b  in  the  (Jc  +  l)st  term  is  I;  and  that  of  a  is  n  -  k. 

(2)  In  the  coefficient  of  any  term  the  last  factor  in  the  denominator 
is  the  same  as  the  exponent  of  b  in  that  term,  and  the  last  factor  in 
the  numerator  is  one  more  than  the  exponent  of  (7. 

Hence  the  (/V  +  l)st  term,  which  is  called  the  general  term  is 

„(„_l)(„_2)(/7-3V--(/7-*  +  l)a„,/,M  n 

1-2-3-4-5-* 


192  THE  BINOMIAL   FORMULA 

224.  The  process  of  writing  out  the  power  of  a  binomial  is 
called  expanding  the  binomial,  and  the  result  is  called  the  ex- 
pansion of  the  binomial. 

Ex.  1.    Expand  (x  —  y)4. 
In  this  case  a  =  x,  b  =  —  y,  n  =  4. 
Hence  substituting  in  formula  I, 

(I  -  vy  =  z>  +  i  x\  -  ,,)  +  ±ii=±>  x-J(>  .,)'-•+  4"~,njit~'-)i-(  -  y)» 
+  4(4-l)(4-2X4-3) 

1  •  o  •  4 

4  13        ,    4-3     o    o         4.3-2        3    ,    4-  3-2.1     4  ,.„ 

Hence    (x  -  y)4  =  .r4  -  4  x3,-/  +  6  .'-//-  -  4  xy*  +  >/*.  (3) 

Notice  that  this  is  precisely  the  same  as  the  expansion  of  (x  +  y)* 
except  that  every  other  term  beginning  with  the  second  is  negative. 

Ex.  2.    Expand  (1-2  y)5. 
Here  0=  1,  b  =  —  2  //,  n  =  .1. 

Since  the  coefficients  in  the  expansion  of  (a  +  h)h  are  1,  5, 10,  10,  •">.  1. 
we  write  at  once, 

(1-2^)6=  l*  +  5.1*.(-2y)  +10.P.(-2y)2 

+  10  ■  l2 •  (-  2  //)3  +  5  •  1  •  (-  2  //)4  +  (-  2  y)5 
=  1  -  10  y  +  40  //-  -  80  y8  +  80  1/  -  32  y6. 

Ex.  3.   Expand  f-+|Y- 

Remembering  the  coefficients  just  given,  we  write  at  once, 

l+!V-(!V+»aYfeUio(!W!V+io'!W« 


r/  \x/    \3/  \./7    \3/  \x/    \3 


"©O'+GD'e+i 


1     :,  v     10 y*    h}>l     *_£  +  jL 

x5      Sx*      9  x3     27  x3      81  x      243" 


EXPANSION   OF  BINOMIALS  193 

In    a    similar    manner  any   positive    integral    power    of    a 
binomial  may  be  written. 

Ex.  4.    Write  the  sixth  term  in  the  expansion  of  (x—  2  y)w 
without  computing  any  other  term. 

From  II,  §  223,  we  know  the  (k  +  l)st  term  for  the  nth  power  of 

a  +  b,  namely,        .       .  OA       ,       ,      1 

n  (n  -  1 )  (  n  -  2)  ■  •  •  ( «  -  k  +  1 )  aH_tfjt 

2.3-4...  k 

In   this   case  a  =  x,   b  =  —  2  >/,   >i  =  10,    k  +  1  =  6.     Hence  k  =  5. 

Substituting  these  particular  values,  we  have 

10(10-l)(10-2)-(10-5  +  l)xl0_5( _  2  ?/)5 
2  •  3  •  4  •  5 

10  •  9  •  8  •  7  •  fi   .,     QO    ~ 

2  •  3  •  4  •  5         ^  J  ' 

=  -  32  -  252  j5//5  =  -  8064  j-5//5. 

EXERCISES 

1.  Make  a  list  of  the  coefficients  for  each  power  of  a  binomial 
from  the  2d  to  the  10th. 

Expand  the  following : 

2.  (x-yf.  9.  (x'-^y.  n     (t_t 

3.  (2x  +  3f.  10-  (ar'  +  ST8)5-  ^      * 

4.  (3*4-2^.         "■  <«-»>'■  18.    ^|-jfV5Y 

12.  r*4-t/V.  V.t  y 


5.    (3+2/)5- 


12.   (oj  +  2/)9. 

1 

19. 


13.    (m-n)»  /V»  ,%ftV 

6.  (.r3  +  2/)6.  14.    (>-*  +  s2)4.  '    \^V       *n 

7.  (^-^)6.  15.    (c--d-^.  ^     /c^_j^ 

8.  (a?-y2)7.  16.    (V«-V&)6.  '    Vvd4 
21.    (2<ftr*-3&y-*)4.           22.    (3  xy-z  -  x~3yy. 


194  THE  BINOMIAL   FORMULA 

In  each  of  the  following  find  the  term  called  for  without 
finding  any  other  term  : 

23.  The  5th  term  of  (a  4-6)12. 

24.  The  7th  term  of  (3  x  —  2y)n. 

25.  The  6th  term  of  (V.v-  \  y)w. 

26.  The  Oth  term  of  (x  -  y  i   . 

27.  The  8th  term  (if  ( \  m  -  \  n  i18. 

28.  The  7th  term  of  (a2b  —  ab     . 

29.  The  (ith  term  of  («— or1)2*. 

30.  The  11th  term  of  {xhj— x~2y-x) 

31.  The  5th  term  from  each  end  of  the  expansion  of  (n  —  h  ,-". 

32.  The  7th  term  from  each  end  of  (a\  a  —  b~\  b)21. 

33.  "Which  term,  counting  from  the  beginning,  has  the  same 
coefficient  as  the  7th  term  of  (a  +  &)10?  Verify  by  finding  both 
coefficients.     How  do  the  exponents  differ  in  these  terms  '.' 

34.  "What  other  term  has  the  same  coefficient  as  the  l'.Mh 
term  of  (a  +  &)24?  How  do  the  exponents  differ  '.'  Find  in  the 
shortest  way  the  21st  term  of  (a  -I-&)25. 

35.  Find  the  87th  term  of  (a  4-  &)90. 
3G.    Find  the  53d  term  of  (a^  —  &^)56. 

37.  What  other  term  lias  the  same  coefficient  as  the  5th 
term  in  flic  expansion  of  U'4-?/)19? 

38.  Expand  [c  4-&)  4-  c]3  by  the  binomial  formula. 

39.  Expand  [1  +  (2  x  +  3  //)]'  by  the  binomial  formula. 

40.  Expand  (2x  —  3y+  iz)s  by  the  binomial  formula. 

41.  Write  the  Qc  + 1) si  term  of  (a  +  b)n.  Write  the  (n+1) si 
term  of  («  +  &)".  Show  that  the  next  and  also  all  succeeding 
terms  after  the  (ra  +  l  isl  term  have  zero  coefficients, thus  prov- 

iii-  I  hat  there  are  exactly  /'  +  1  terras  in  the  expansion. 


MATHEMATICS 


Introduction  to  Geometry 

By  William  SCHOCH,  of  the  Crane  Manual  Training  High  School, 
Chicago.     121110,  cloth,  142  pages.     Price,  60  cents. 

THIS  book  is  intended  for  pupils  in  the  upper  grades  of  the 
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Plane  Figures,  Parallel  Lines,  Triangles,  etc.)  by  making  use 
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problems  are  given  and  the  pupil's  knowledge  is  constantly 
tested  by  his  power  to  apply  it. 

Principles  of  Plane  Geometry 

By  J.  W.  MacDonald,  Agent  of  the  Massachusetts  Board  of  Educa- 
tion.    i6mo,  paper,  70  pages.     Price,  30  cents. 

Logarithmic  and  Other  Mathematical  Tables 

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ford Junior  University,  California.    8vo,  cloth,  148  pages.     Price,  $1.00. 

VARIOUS  mechanical  devices  make  this  work  specially  easy 
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to  the  eye,  as  a  piece  of  careful  and  successful  book-making. 

Plane  Trigonometry 

By  Professor  R.  D.  Bohannan,  of  the  State  University,  Columbus, 
Ohio.     Cloth,  379  pages.     Price,  $2.50. 


SOME  of  the  feat 
ing  variety  of 


OME  of  the  features  of  this  book  are  the  use  of  triangles  hav- 
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have  if  they  had  been  actually  measured  in  the  field  ;  the  sugges- 
tion of  simple  laboratory  exercises  ;  problems  from  related  sub- 
jects, such  as  Physics,  Analytical  Geometry,  and  Surveying. 


MATHEMATICS 


High   School  Algebra:    Elementary  Course 

By  H.  E.  SLAUGHT,  Assistant  Professor  of  Mathematics  in  the  Univer- 
sity of  Chicago,  and  X.  ].  LENNES,  Instructor  in  Mathematics  in  the 
Massachusetts  Institute  of  Technology.  i2mo,  cloth,  309  pages. 
Price,  $1.00. 

THIS  book  embodies  the  methods  of  what  might  be  called  the 
new  school  of  Algebra  teaching,  as  revealed  in  numerous 
recent  discussions  and  articles. 

Some  of  the  important  features  of  the  Elementary  Course  are  :  — 

I.  Algebra  is  vitally  and  persistently  connected  with  Arithme- 
tic. Each  principle  in  the  book  is  first  studied  in  its  application 
to  numbers  in  the  Arabic  notation.  Principles  of  Algebra  are 
thus  connected  with  those  already  known  in  Arithmetic. 

II.  The  principles  of  Algebra  are  enunciated  in  a  small  number 
of  easy  statements,  eighteen  in  all.  The  purpose  of  these  prin- 
ciples is  to  arrange  in  simple  form  a  codification  of  those  opera- 
tions of  Algebra  that  are  sufficiently  different  from  the  ordinary 
operations  of  elementary  Arithmetic  to  require  special  emphasis. 

III.  The  main  purpose  of  the  book  is  the  solution  of  problems 
rather  than  the  construction  of  a  purely  theoretical  doctrine  as  an 
end  in  itself.  An  attempt  is  made  to  connect  each  principle  in  a 
vital  manner  with  the  learner's  experience  by  using  it  in  the  solu- 
tion of  a  large  number  of  simple  problems. 

IV.  The  descriptive  problems,  as  distinguished  from  the  mere 
examples,  are  over  seven  hundred  in  number,  about  twice  as 
many  as  are  contained  in  any  other  book.  A  large  share  of  these 
problems  deal  with  simple  geometrical  and  physical  data.  Many 
others  contain  real  data,  so  that  the  pupil  feels  that  he  is  deal- 
ing with  problems  connected  with  life.  The  authors  believe 
that  it  is  possible  to  make  problems  so  interesting  as  to  answer 
the  ordinary  schoolboy  query — "What  is  Algebra  ^ood  for?11 

V.  The  order  of  topics  has  been  changed  from  the  traditional 
one,  so  that  those  subjects  will  come  first  which  have  a  direct 
bearing  on  the  solution  of  problems,  which  is  looked  upon  as  the 
first  purpose  of  Algebra. 


MATHEMATICS 


High   School  Algebra:    Advanced  Course 

By  H.  E.  Slaught,  of  the  University  of  Chicago,  and  N.  J.  Lennes, 
of  the  Massachusetts  Institute  of  Technology.  i2mo,  cloth,  202  pages. 
Price,  75  cents. 

THE  Advanced  Course  contains  a  complete  review  of  the  sub- 
jects treated  in  the  Elementary  Course,  together  with  enough 
additional  topics  to  meet  the  requirements  of  the  most  exacting 
technical  and  scientific  schools. 

The  subjects  already  familiar  from  use  of  the  Elementary  Course 
are  treated  from  a  more  mature  standpoint.  The  theorems  are 
framed  from  a  definite  body  of.  axioms.  The  main  purposes  of 
this  part  of  the  book  are:  (a)  To  study  the  logical  aspects  of 
elementary  Algebra;  (b)  to  give  the  needed  drill  in  performing 
algebraic  operations.  To  this  end  the  exercises  are  considerably 
more  complicated  than  in  the  Elementary  Course. 

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mentary Course  are  worked  out  in  greater  detail.  The  chapters 
on  Quadratics  and  Radicals  contain  rich  sets  of  examples  based 
largely  upon  a  few  geometrical  relations  which  the  pupils  used 
freely  in  the  grammar  schools.  The  chapters  of  Ratio,  Propor- 
tion, and  Progression  also  contain  extensive  sets  of  problems. 

The  chapters  are : 

I.    Fundamental  Laws.  VII.    Quadratics. 

II.    Fundamental  Operations.  VIII.    Algebraic  Fractions. 

III.  Integral    Equations    of    the  IX.    Ratio,  Proportion,  and  Vari- 

First  Degree.  ation. 

IV.  Integral  Linear  Equations  in  X.    Exponents  and  Radicals. 

Two  or  More  Variables.  XL    Progressions. 

V.    Factoring.  XII.    Logarithms. 

VI.    Powers  and  Roots.  XIII.    Binomial  Theorem. 

A  Primary  Algebra 

By  J.  W.  MacDonald,  Agent  of  the  Massachusetts  Board  of  Educa- 
tion.    i6mo,  cloth.     The  Student's  Manual,  92  pages.     Price,  30  cents. 
The  Complete  Edition,  for  teachers,  21S  pages.     Price,  75  cents. 
69 


MATIIKMATICS 


Elements  of  Algebra 

By  Professor  James  M.  Taylor,  Colgate  University,  Hamilton,  N.Y. 
i2mo,  half  leather,  461  pages.     Price,  $1.12. 

IN  the  Elements  of  Algebra  the  author  has  aimed  to  give  a 
treatment  of  the  subject  so  simple  that  it  may  be  used  by 
beginners,  and  at  the  same  time  so  scientific  and  logical  as  to 
leave  nothing  to  be  unlearned  as  progress  is  made  in  the  study 
of  mathematics.  Simplicity  is  not  attained  by  the  use  of  inexact 
statements  and  mechanical  methods,  but  by  giving  demonstra- 
tions in  full  and  making  explanations  as  concrete  as  possible. 

General  principles  are  first  illustrated  by  particular  examples. 
Each  new  principle  is  demonstrated  from  laws  that  have  preceded  it. 

Topics  which  have  unique  treatment  or  special  emphasis  are — 
the  meaning  and  advantages  of  the  literal  notation  ;  distinction 
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and  systems  ;  the  graph  ;  theory  of  limits  ;  imaginary  numbers  ; 
the  remainder  theorem. 

An  Academic  Algebra 

By   Professor  J.    M.    TAYLOR,    Colgate   University,    Hamilton,    N.Y. 
i6mo,  cloth,  348  pages.     Price,  $1.00. 

THIS  book  is  distinguished  by  its  scientific  and  logical 
method.  The  author  believes  that  there  is  economy  of  time 
in  teaching  the  fundamental  theory  to  the  pupil  at  once  and  in  a 
form  which  he  will  not  need  to  unlearn  when  he  proceeds  to  higher 
mathematics. 

The  equation  is  introduced  as  early  as  possible  and  the  bear- 
ing of  each  succeeding  subject  on  the  solution  of  the  equation 
is  made  clear. 

Each  principle  is  proved  once  for  all  whatever  form  it  may 
afterwards  assume. 

Factoring  is  made  fundamental  in  the  solution  of  quadratics. 
Equivalent   equations    and    systems    of  equations    are    clearly 
proved,  fully  illustrated,  and  amply  applied. 

70 


mathematics 


A  College  Algebra 

By    Professor  J.    M.   Taylor,   Colgate   University,   Hamilton,    N.Y. 
i6mo,  cloth,  373  pages.     Price,  $1.50. 

THIS  book  is  very  much  smaller  than  the  ordinary  college 
Algebra  though  covering  practically  the  same  ground.  The 
reason  for  this  is  that  less  space  is  devoted  to  the  first  part, 
which  is  a  review  of  preparatory  Algebra. 

In  the  First  Part  is  found  an  outline  of  the  fundamental  prin- 
ciples of  Algebra  usually  required  for  admission  to  a  college  or 
scientific  school.  The  subjects  of  Equivalent  Equations  and 
Equivalent  Systems  of  Equations  are  given  especial  emphasis. 

In  the  Second  Part  a  full  discussion  of  the  Theory  of  Limits  is 
followed  by  one  of  its  most  important  applications.  Differentia- 
tion, leading  to  clear  and  concise  proofs  of  the  Binomial  Theo- 
rem, Logarithmic  Series,  and  Exponential  Series,  as  particular 
cases  of  Maclaurin's  Formula.  This  affords  the  student  an 
easy  introduction  to  the  concepts  of  higher  mathematics. 

The  order  of  chapters  may  be  varied,  and  subjects  that  are 
recommended  for  a  second  reading,  like  Summation  of  Series  and 
Continued  Fractions,  are  marked  by  an  asterisk. 

There  are  chapters  on  Determinants,  and  on  the  Graphic  Solu- 
tion of  Equations  and  Systems  of  Equations. 

This  book  is  printed  both  with  and  without  the  answers. 

Calculus  with  Applications 

By  Eli.en  Hayes,  Professor  of  Mathematics  at  Wellesley  College. 
121110,  cloth,  170  pages.     Price,  $1.20. 

THIS  book  is  a  reading  lesson  in  applied  mathematics,  in- 
tended for  persons  who  wish,  without  taking  long  courses 
in  mathematics,  to  know  what  the  calculus  is  and  how  to  use  it, 
either  as  applied  to  other  sciences,  or  for  purposes  of  general  cul- 
ture. Nothing  is  included  in  the  book  that  is  not  a  means  to 
this  end.  All  fancy  exercises  are  avoided,  and  the  problems  are 
for  the  most  part  real  ones  from  mechanics  or  astronomy. 

71 


MATHEMATICS 


Plane  and   Spherical  Trigonometry 

By  President  Elmer  A.  LYMAN,  Michigan  State  Normal  College,  and 
Professor  Edwin  C.  Goddard,  University  of  Michigan.  Cloth,  149 
pages.     Price,  90  cents. 

MANY  American  text-books  on  Trigonometry  treat  the  solu- 
tion of  triangles  fully ;  English  text-books  elaborate  ana- 
lytical Trigonometry.  No  book  seems  to  have  met  both  needs 
adequately.  It  is  the  aim  of  Lyman  and  Goddard  to  find  the 
golden  mean  between  the  preponderance  of  practice  in  the  one 
and  that  of  theory  in  the  other. 

The  book  is  a  direct  outgrowth  of  the  classroom  experience  of 
the  authors,  and  it  has  no  great  resemblance  to  the  traditional  text- 
book in  Trigonometry.  Two  subjects  that  have  called  for  origi- 
nal treatment  or  emphasis  are  the  proof  of  the  formulae  for  the 
functions  of  a  ±  /3  and  inverse  functions.  By  dint  of  much  prac- 
tice extended  over  as  long  a  time  as  possible  it  is  hoped  to  give 
the  pupil  a  command  of  logarithms  that  will  stay. 

Computation  Tables 

By  I.V.MAX  and  GODDARD.     Price,  50  cents. 

THESE  are  five-place  logarithmic  tables  which  have  been 
prepared  to  accompany  Lyman  and  Goddard's  Trigonome- 
trv.  They  have  been  rendered  exceedingly  convenient  by  the 
use  of  distinctive  type,  which  makes  it  easy  to  find  the  numbers 
sought. 

The  natural  and  logarithmic  series  are  given  side  by  side  in 
the  tables.  The  column  of  difference  is  in  unusually  direct 
proximity  to  the  logarithms  themselves. 

Plane  and  Spherical  Trigonometry,  with  Compu- 
tation Tables 


By  Lyman  and  Goddard.    Complete   Edition.     Cloth,    214  piges. 

si. 20. 


RETURN     CIRCULAR  ubl2^_i 

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THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


